Abstract Algebra: Applications to Galois Theory, Algebraic Geometry and Cryptography

De Gruyter Graduate

Celine Carstensen Benjamin Fine Gerhard Rosenberger

Abstract Algebra Applications to Galois Theory, Algebraic Geometry and Cryptography

De Gruyter

Mathematics Subject Classification 2010: Primary: 12-01, 13-01, 16-01, 20-01; Secondary: 01-01, 08-01, 11-01, 14-01, 94-01.

This book is Volume 11 of the Sigma Series in Pure Mathematics, Heldermann Verlag.

ISBN 978-3-11-025008-4 e-ISBN 978-3-11-025009-1 Library of Congress Cataloging-in-Publication Data Carstensen, Celine. Abstract algebra : applications to Galois theory, algebraic geometry, and cryptography / by Celine Carstensen, Benjamin Fine, and Gerhard Rosenberger. p. cm. ⫺ (Sigma series in pure mathematics ; 11) Includes bibliographical references and index. ISBN 978-3-11-025008-4 (alk. paper) 1. Algebra, Abstract. 2. Galois theory. 3. Geometry, Algebraic. 4. Crytography. I. Fine, Benjamin, 1948⫺ II. Rosenberger, Gerhard. III. Title. QA162.C375 2011 5151.02⫺dc22 2010038153

Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available in the Internet at http://dnb.d-nb.de. ” 2011 Walter de Gruyter GmbH & Co. KG, Berlin/New York Typesetting: Da-TeX Gerd Blumenstein, Leipzig, www.da-tex.de Printing and binding: AZ Druck und Datentechnik GmbH, Kempten ⬁ Printed on acid-free paper Printed in Germany www.degruyter.com

Preface

Traditionally, mathematics has been separated into three main areas; algebra, analysis and geometry. Of course there is a great deal of overlap between these areas. For example, topology, which is geometric in nature, owes its origins and problems as much to analysis as to geometry. Further the basic techniques in studying topology are predominantly algebraic. In general, algebraic methods and symbolism pervade all of mathematics and it is essential for anyone learning any advanced mathematics to be familiar with the concepts and methods in abstract algebra. This is an introductory text on abstract algebra. It grew out of courses given to advanced undergraduates and beginning graduate students in the United States and to mathematics students and teachers in Germany. We assume that the students are familiar with Calculus and with some linear algebra, primarily matrix algebra and the basic concepts of vector spaces, bases and dimensions. All other necessary material is introduced and explained in the book. We assume however that the students have some, but not a great deal, of mathematical sophistication. Our experience is that the material in this can be completed in a full years course. We presented the material sequentially so that polynomials and field extensions preceded an in depth look at group theory. We feel that a student who goes through the material in these notes will attain a solid background in abstract algebra and be able to move on to more advanced topics. The centerpiece of these notes is the development of Galois theory and its important applications, especially the insolvability of the quintic. After introducing the basic algebraic structures, groups, rings and fields, we begin the theory of polynomials and polynomial equations over fields. We then develop the main ideas of field extensions and adjoining elements to fields. After this we present the necessary material from group theory needed to complete both the insolvability of the quintic and solvability by radicals in general. Hence the middle part of the book, Chapters 9 through 14 are concerned with group theory including permutation groups, solvable groups, abelian groups and group actions. Chapter 14 is somewhat off to the side of the main theme of the book. Here we give a brief introduction to free groups, group presentations and combinatorial group theory. With the group theory material in hand we return to Galois theory and study general normal and separable extensions and the fundamental theorem of Galois theory. Using this we present several major applications of the theory including solvability by radicals and the insolvability of the quintic, the fundamental theorem of algebra, the construction of regular n-gons and the famous impossibilities; squaring the circling, doubling the cube and trisecting an angle. We

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Preface

finish in a slightly different direction giving an introduction to algebraic and group based cryptography. October 2010

Celine Carstensen Benjamin Fine Gerhard Rosenberger

Contents

Preface 1 Groups, Rings and Fields 1.1 Abstract Algebra . . . . . . . . . . . . 1.2 Rings . . . . . . . . . . . . . . . . . . 1.3 Integral Domains and Fields . . . . . . 1.4 Subrings and Ideals . . . . . . . . . . . 1.5 Factor Rings and Ring Homomorphisms 1.6 Fields of Fractions . . . . . . . . . . . 1.7 Characteristic and Prime Rings . . . . . 1.8 Groups . . . . . . . . . . . . . . . . . . 1.9 Exercises . . . . . . . . . . . . . . . .

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1 1 2 4 6 9 13 14 17 19

2 Maximal and Prime Ideals 2.1 Maximal and Prime Ideals . . . . . . . . . . 2.2 Prime Ideals and Integral Domains . . . . . . 2.3 Maximal Ideals and Fields . . . . . . . . . . 2.4 The Existence of Maximal Ideals . . . . . . . 2.5 Principal Ideals and Principal Ideal Domains . 2.6 Exercises . . . . . . . . . . . . . . . . . . .

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3 Prime Elements and Unique Factorization Domains 3.1 The Fundamental Theorem of Arithmetic . . . . 3.2 Prime Elements, Units and Irreducibles . . . . . 3.3 Unique Factorization Domains . . . . . . . . . . 3.4 Principal Ideal Domains and Unique Factorization 3.5 Euclidean Domains . . . . . . . . . . . . . . . . 3.6 Overview of Integral Domains . . . . . . . . . . 3.7 Exercises . . . . . . . . . . . . . . . . . . . . .

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4 Polynomials and Polynomial Rings 4.1 Polynomials and Polynomial Rings . . . . . . . . . . . 4.2 Polynomial Rings over Fields . . . . . . . . . . . . . . 4.3 Polynomial Rings over Integral Domains . . . . . . . . 4.4 Polynomial Rings over Unique Factorization Domains 4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . .

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Contents

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66 66 69 70 74 75 78

6 Field Extensions and Compass and Straightedge Constructions 6.1 Geometric Constructions . . . . . . . . . . . . . . . . . . . 6.2 Constructible Numbers and Field Extensions . . . . . . . . . 6.3 Four Classical Construction Problems . . . . . . . . . . . . 6.3.1 Squaring the Circle . . . . . . . . . . . . . . . . . . 6.3.2 The Doubling of the Cube . . . . . . . . . . . . . . 6.3.3 The Trisection of an Angle . . . . . . . . . . . . . . 6.3.4 Construction of a Regular n-Gon . . . . . . . . . . . 6.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .

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80 80 80 83 83 83 83 84 89

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91 91 94 100 100 101 105 109 111

8 Splitting Fields and Normal Extensions 8.1 Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Normal Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

113 113 115 118

9 Groups, Subgroups and Examples 9.1 Groups, Subgroups and Isomorphisms 9.2 Examples of Groups . . . . . . . . . 9.3 Permutation Groups . . . . . . . . . . 9.4 Cosets and Lagrange’s Theorem . . . 9.5 Generators and Cyclic Groups . . . . 9.6 Exercises . . . . . . . . . . . . . . .

119 119 121 125 128 133 139

7

Field Extensions 5.1 Extension Fields and Finite Extensions . . . . 5.2 Finite and Algebraic Extensions . . . . . . . 5.3 Minimal Polynomials and Simple Extensions 5.4 Algebraic Closures . . . . . . . . . . . . . . 5.5 Algebraic and Transcendental Numbers . . . 5.6 Exercises . . . . . . . . . . . . . . . . . . .

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Kronecker’s Theorem and Algebraic Closures 7.1 Kronecker’s Theorem . . . . . . . . . . . . . . . . . . 7.2 Algebraic Closures and Algebraically Closed Fields . . 7.3 The Fundamental Theorem of Algebra . . . . . . . . . 7.3.1 Splitting Fields . . . . . . . . . . . . . . . . . 7.3.2 Permutations and Symmetric Polynomials . . . 7.4 The Fundamental Theorem of Algebra . . . . . . . . . 7.5 The Fundamental Theorem of Symmetric Polynomials 7.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . .

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Contents

10 Normal Subgroups, Factor Groups and Direct Products 10.1 Normal Subgroups and Factor Groups . . . . . . . . 10.2 The Group Isomorphism Theorems . . . . . . . . . . 10.3 Direct Products of Groups . . . . . . . . . . . . . . 10.4 Finite Abelian Groups . . . . . . . . . . . . . . . . . 10.5 Some Properties of Finite Groups . . . . . . . . . . . 10.6 Exercises . . . . . . . . . . . . . . . . . . . . . . .

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141 141 146 149 151 156 160

11 Symmetric and Alternating Groups 11.1 Symmetric Groups and Cycle Decomposition 11.2 Parity and the Alternating Groups . . . . . . 11.3 Conjugation in Sn . . . . . . . . . . . . . . . 11.4 The Simplicity of An . . . . . . . . . . . . . 11.5 Exercises . . . . . . . . . . . . . . . . . . .

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161 161 164 167 168 170

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12 Solvable Groups 12.1 Solvability and Solvable Groups . . . . . . . . . . . 12.2 Solvable Groups . . . . . . . . . . . . . . . . . . . . 12.3 The Derived Series . . . . . . . . . . . . . . . . . . 12.4 Composition Series and the Jordan–Hölder Theorem 12.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . 13 Groups Actions and the Sylow Theorems 13.1 Group Actions . . . . . . . . . . . . . . . . 13.2 Conjugacy Classes and the Class Equation . 13.3 The Sylow Theorems . . . . . . . . . . . . 13.4 Some Applications of the Sylow Theorems 13.5 Exercises . . . . . . . . . . . . . . . . . .

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180 180 181 183 187 191

14 Free Groups and Group Presentations 14.1 Group Presentations and Combinatorial Group Theory 14.2 Free Groups . . . . . . . . . . . . . . . . . . . . . . . 14.3 Group Presentations . . . . . . . . . . . . . . . . . . . 14.3.1 The Modular Group . . . . . . . . . . . . . . 14.4 Presentations of Subgroups . . . . . . . . . . . . . . . 14.5 Geometric Interpretation . . . . . . . . . . . . . . . . 14.6 Presentations of Factor Groups . . . . . . . . . . . . . 14.7 Group Presentations and Decision Problems . . . . . . 14.8 Group Amalgams: Free Products and Direct Products . 14.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . .

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192 192 193 198 200 207 209 212 213 214 216

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Contents

15 Finite Galois Extensions 15.1 Galois Theory and the Solvability of Polynomial Equations 15.2 Automorphism Groups of Field Extensions . . . . . . . . 15.3 Finite Galois Extensions . . . . . . . . . . . . . . . . . . 15.4 The Fundamental Theorem of Galois Theory . . . . . . . 15.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .

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217 217 218 220 221 231

16 Separable Field Extensions 16.1 Separability of Fields and Polynomials 16.2 Perfect Fields . . . . . . . . . . . . . 16.3 Finite Fields . . . . . . . . . . . . . . 16.4 Separable Extensions . . . . . . . . . 16.5 Separability and Galois Extensions . . 16.6 The Primitive Element Theorem . . . 16.7 Exercises . . . . . . . . . . . . . . .

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233 233 234 236 238 241 245 247

17 Applications of Galois Theory 17.1 Applications of Galois Theory . . . . 17.2 Field Extensions by Radicals . . . . . 17.3 Cyclotomic Extensions . . . . . . . . 17.4 Solvability and Galois Extensions . . 17.5 The Insolvability of the Quintic . . . . 17.6 Constructibility of Regular n-Gons . . 17.7 The Fundamental Theorem of Algebra 17.8 Exercises . . . . . . . . . . . . . . .

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248 248 248 252 253 254 259 261 263

18 The Theory of Modules 18.1 Modules Over Rings . . . . . . . . . . . . . . . . . . . . 18.2 Annihilators and Torsion . . . . . . . . . . . . . . . . . . 18.3 Direct Products and Direct Sums of Modules . . . . . . . 18.4 Free Modules . . . . . . . . . . . . . . . . . . . . . . . . 18.5 Modules over Principal Ideal Domains . . . . . . . . . . . 18.6 The Fundamental Theorem for Finitely Generated Modules 18.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .

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285 285 286 288 294

19 Finitely Generated Abelian Groups 19.1 Finite Abelian Groups . . . . . . . . . . . . . . . . . 19.2 The Fundamental Theorem: p-Primary Components 19.3 The Fundamental Theorem: Elementary Divisors . . 19.4 Exercises . . . . . . . . . . . . . . . . . . . . . . .

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Contents

20 Integral and Transcendental Extensions 20.1 The Ring of Algebraic Integers . . . 20.2 Integral ring extensions . . . . . . . 20.3 Transcendental field extensions . . . 20.4 The transcendence of e and  . . . . 20.5 Exercises . . . . . . . . . . . . . .

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295 295 298 302 307 310

21 The Hilbert Basis Theorem and the Nullstellensatz 21.1 Algebraic Geometry . . . . . . . . . . . . . . . . . . . 21.2 Algebraic Varieties and Radicals . . . . . . . . . . . . 21.3 The Hilbert Basis Theorem . . . . . . . . . . . . . . . 21.4 The Hilbert Nullstellensatz . . . . . . . . . . . . . . . 21.5 Applications and Consequences of Hilbert’s Theorems 21.6 Dimensions . . . . . . . . . . . . . . . . . . . . . . . 21.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . .

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312 312 312 314 315 317 320 325

22 Algebraic Cryptography 22.1 Basic Cryptography . . . . . . . . . . . . . . . . . 22.2 Encryption and Number Theory . . . . . . . . . . 22.3 Public Key Cryptography . . . . . . . . . . . . . . 22.3.1 The Diffie–Hellman Protocol . . . . . . . . 22.3.2 The RSA Algorithm . . . . . . . . . . . . 22.3.3 The El-Gamal Protocol . . . . . . . . . . . 22.3.4 Elliptic Curves and Elliptic Curve Methods 22.4 Noncommutative Group based Cryptography . . . 22.4.1 Free Group Cryptosystems . . . . . . . . . 22.5 Ko–Lee and Anshel–Anshel–Goldfeld Methods . . 22.5.1 The Ko–Lee Protocol . . . . . . . . . . . . 22.5.2 The Anshel–Anshel–Goldfeld Protocol . . 22.6 Platform Groups and Braid Group Cryptography . 22.7 Exercises . . . . . . . . . . . . . . . . . . . . . .

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326 326 331 335 336 337 339 341 342 345 349 350 350 351 356

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Bibliography

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Index

363

Chapter 1

Groups, Rings and Fields

1.1

Abstract Algebra

Abstract algebra or modern algebra can be best described as the theory of algebraic structures. Briefly, an algebraic structure is a set S together with one or more binary operations on it satisfying axioms governing the operations. There are many algebraic structures but the most commonly studied structures are groups, rings, fields and vector spaces. Also widely used are modules and algebras. In this first chapter we will look at some basic preliminaries concerning groups, rings and fields. We will only briefly touch on groups here, a more extensive treatment will be done later in the book. Mathematics traditionally has been subdivided into three main areas – analysis, algebra and geometry. These areas overlap in many places so that it is often difficult to determine whether a topic is one in geometry say or in analysis. Algebra and algebraic methods permeate all these disciplines and most of mathematics has been algebraicized – that is uses the methods and language of algebra. Groups, rings and fields play a major role in the modern study of analysis, topology, geometry and even applied mathematics. We will see these connections in examples throughout the book. Abstract algebra has its origins in two main areas and questions that arose in these areas – the theory of numbers and the theory of equations. The theory of numbers deals with the properties of the basic number systems – integers, rationals and reals while the theory of equations, as the name indicates, deals with solving equations, in particular polynomial equations. Both are subjects that date back to classical times. A whole section of Euclid’s elements is dedicated to number theory. The foundations for the modern study of number theory were laid by Fermat in the 1600s and then by Gauss in the 1800s. In an attempt to prove Fermat’s big theorem Gauss introduced the complex integers a C bi where a and b are integers and showed that this set has unique factorization. These ideas were extended by Dedekind and Kronecker who developed a wide ranging theory of algebraic number fields and algebraic integers. A large portion of the terminology used in abstract algebra, rings, ideals, factorization comes from the study of algebraic number fields. This has evolved into the modern discipline of algebraic number theory. The second origin of modern abstract algebra was the problem of trying to determine a formula for finding the solutions in terms of radicals of a fifth degree polynomial. It was proved first by Ruffini in 1800 and then by Abel that it is impossible to find a formula in terms of radicals for such a solution. Galois in 1820 extended

2

Chapter 1 Groups, Rings and Fields

this and showed that such a formula is impossible for any degree five or greater. In proving this he laid the groundwork for much of the development of modern abstract algebra especially field theory and finite group theory. Earlier, in 1800, Gauss proved the fundamental theorem of algebra which says that any nonconstant complex polynomial equation must have a solution. One of the goals of this book is to present a comprehensive treatment of Galois theory and a proof of the results mentioned above. The locus of real points .x; y/ which satisfy a polynomial equation f .x; y/ D 0 is called an algebraic plane curve. Algebraic geometry deals with the study of algebraic plane curves and extensions to loci in a higher number of variables. Algebraic geometry is intricately tied to abstract algebra and especially commutative algebra. We will touch on this in the book also. Finally linear algebra, although a part of abstract algebra, arose in a somewhat different context. Historically it grew out of the study of solution sets of systems of linear equations and the study of the geometry of real n-dimensional spaces. It began to be developed formally in the early 1800s with work of Jordan and Gauss and then later in the century by Cayley, Hamilton and Sylvester.

1.2

Rings

The primary motivating examples for algebraic structures are the basic number systems; the integers Z, the rational numbers Q, the real numbers R and the complex numbers C. Each of these has two basic operations, addition and multiplication and form what is called a ring. We formally define this. Definition 1.2.1. A ring is a set R with two binary operations defined on it, addition, denoted by C, and multiplication, denoted by  , or just by juxtaposition, satisfying the following six axioms: (1) Addition is commutative: a C b D b C a for each pair a; b in R. (2) Addition is associative: a C .b C c/ D .a C b/ C c for a; b; c 2 R. (3) There exists an additive identity, denoted by 0, such that a C 0 D a for each a 2 R. (4) For each a 2 R there exists an additive inverse, denoted by a, such that a C .a/ D 0. (5) Multiplication is associative: a.bc/ D .ab/c for a; b; c 2 R. (6) Multiplication is left and right distributive over addition: a.b C c/ D ab C ac and .b C c/a D ba C ca for a; b; c 2 R.

3

Section 1.2 Rings

If in addition (7) Multiplication is commutative: ab D ba for each pair a; b in R. then R is a commutative ring. Further if (8) There exists a multiplicative identity denoted by 1 such that a  1 D a and 1  a D a for each a in R. then R is a ring with identity. If R satisfies (1) through (8) then R is a commutative ring with an identity. A set G with one operation, C, on it satisfying axioms (1) through (4) is called an abelian group. We will discuss these further later in the chapter. The numbers systems Z; Q; R; C are all commutative rings with identity. A ring R with only one element is called trivial. A ring R with identity is trivial if and only if 0 D 1. A finite ring is a ring R with only finitely many elements in it. Otherwise R is an infinite ring. Z; Q; R; C are all infinite rings. Examples of finite rings are given by the integers modulo n, Zn , with n > 1. The ring Zn consists of the elements 0; 1; 2; : : : ; n1 with addition and multiplication done modulo n. That is, for example 4  3 D 12 D 2 modulo 5. Hence in Z5 we have 4  3 D 2. The rings Zn are all finite commutative rings with identity. To give examples of rings without an identity consider the set nZ D ¹nz W z 2 Zº consisting of all multiples of the fixed integer n. It is an easy verification (see exercises) that this forms a ring under the same addition and multiplication as in Z but that there is no identity for multiplication. Hence for each n 2 Z with n > 1 we get an infinite commutative ring without an identity. To obtain examples of noncommutative rings we consider matrices. Let M2 .Z/ be the set of 2  2 matrices with integral entries. Addition of matrices is done componentwise, that is       a2 b2 a1 C a2 b1 C b2 a1 b1 C D c1 d1 c2 d2 c1 C c2 d1 C d2 while multiplication is matrix multiplication 

a 1 b1 c1 d1

     a2 b2 a a C b1 c2 a1 b2 C b1 d2  D 1 2 : c2 d 2 c1 a2 C d1 c2 c1 b2 C d1 d2

Then again it is an easy verification (see exercises) that M2 .Z/ forms a ring. Further since matrix multiplication is noncommutative this forms a noncommutative ring. However the identity matrix does form a multiplicative identity for it. M2 .nZ/ with n > 1 provides an example of an infinite noncommutative ring without an identity. Finally M2 .Zn / for n > 1 will give an example of a finite noncommutative ring.

4

1.3

Chapter 1 Groups, Rings and Fields

Integral Domains and Fields

Our basic number systems have the property that if ab D 0 then either a D 0 or b D 0. However this is not necessarily true in the modular rings. For example 2  3 D 0 in Z6 . Definition 1.3.1. A zero divisor in a ring R is an element a 2 R with a ¤ 0 such that there exists an element b ¤ 0 with ab D 0. A commutative ring with an identity 1 ¤ 0 and with no zero divisors is called an integral domain. Notice that having no zero divisors is equivalent to the fact that if ab D 0 in R then either a D 0 or b D 0. Hence Z; Q; R; C are all integral domains but from the example above Z6 is not. In general we have the following. Theorem 1.3.2. Zn is an integral domain if and only if n is a prime. Proof. First of all notice that under multiplication modulo n an element m is 0 if and only if n divides m. We will make this precise shortly. Recall further Euclid’s lemma which says that if a prime p divides a product ab then p divides a or p divides b. Now suppose that n is a prime and ab D 0 in Zn . Then n divides ab. From Euclid’s lemma it follows that n divides a or n divides b. In the first case a D 0 in Zn while in the second b D 0 in Zn . It follows that there are no zero divisors in Zn and since Zn is a commutative ring with an identity it is an integral domain. Conversely suppose Zn is an integral domain. Suppose that n is not prime. Then n D ab with 1 < a < n, 1 < b < n. It follows that ab D 0 in Zn with neither a nor b being zero. Therefore they are zero divisors which is a contradiction. Hence n must be prime. In Q every nonzero element has a multiplicative inverse. This is not true in Z where only the elements 1; 1 have multiplicative inverses within Z. Definition 1.3.3. A unit in a ring R with identity is an element a which has a multiplicative inverse, that is an element b such that ab D ba D 1. If a is a unit in R we denote its inverse by a1 . Hence every nonzero element of Q and of R and of C is a unit but in Z the only units are ˙1. In M2 .R/ the units are precisely those matrices that have nonzero determinant while in M2 .Z/ the units are those integral matrices that have determinant ˙1. Definition 1.3.4. A field F is a commutative ring with an identity 1 ¤ 0 where every nonzero element is a unit. The rationals Q, the reals R and the complexes C are all fields. If we relax the commutativity requirement and just require that in the ring R with identity each nonzero element is a unit then we get a skew field or division ring.

5

Section 1.3 Integral Domains and Fields

Lemma 1.3.5. If F is a field then F is an integral domain. Proof. Since a field F is already a commutative ring with an identity we must only show that there are no zero divisors in F . Suppose that ab D 0 with a ¤ 0. Since F is a field and a is nonzero it has an inverse a1 . Hence a1 .ab/ D a1 0 D 0 H) .a1 a/b D 0 H) b D 0: Therefore F has no zero divisors and must be an integral domain. Recall that Zn was an integral domain only when n was a prime. This turns out to also be necessary and sufficient for Zn to be a field. Theorem 1.3.6. Zn is a field if and only if n is a prime. Proof. First suppose that Zn is a field. Then from Lemma 1.3.5 it is an integral domain, so from Theorem 1.3.2 n must be a prime. Conversely suppose that n is a prime. We must show that Zn is a field. Since we already know that Zn is an integral domain we must only show that each nonzero element of Zn is a unit. Here we need some elementary facts from number theory. If a; b are integers we use the notation ajb to indicate that a divides b. Recall that given nonzero integers a; b their greatest common divisor or GCD d > 0 is a positive integer which is a common divisor, that is d ja and d jb, and if d1 is any other common divisor then d1 jd . We denote the greatest common divisor of a; b by either gcd.a; b/ or .a; b/. It can be proved that given nonzero integers a; b their GCD exists, is unique and can be characterized as the least positive linear combination of a and b. If the GCD of a and b is 1 then we say that a and b are relatively prime or coprime. This is equivalent to being able to express 1 as a linear combination of a and b. Now let a 2 Zn with n prime and a ¤ 0. Since a ¤ 0 we have that n does not divide a. Since n is prime it follows that a and n must be relatively prime, .a; n/ D 1. From the number theoretic remarks above we then have that there exist x; y with ax C ny D 1: However in Zn the element ny D 0 and so in Zn we have ax D 1:

6

Chapter 1 Groups, Rings and Fields

Therefore a has a multiplicative inverse in Zn and is hence a unit. Since a was an arbitrary nonzero element we conclude that Zn is a field. The theorem above is actually a special case of a more general result from which Theorem 1.3.6 could also be obtained. Theorem 1.3.7. Each finite integral domain is a field. Proof. Let F be a finite integral domain. We must show that F is a field. It is clearly sufficient to show that each nonzero element of F is a unit. Let ¹0; 1; r1 ; : : : ; rn º be the elements of F . Let ri be a fixed nonzero element and multiply each element of F by ri on the left. Now if ri rj D ri rk then ri .rj  rk / D 0: Since ri ¤ 0 it follows that rj  rk D 0 or rj D rk . Therefore all the products ri rj are distinct. Hence R D ¹0; 1; r1 ; : : : ; rn º D ri R D ¹0; ri ; ri r1 ; : : : ; ri rn º: Hence the identity element 1 must be in the right-hand list, that is there is an rj such that ri rj D 1. Therefore ri has a multiplicative inverse and is hence a unit. Therefore F is a field.

1.4

Subrings and Ideals

A very important concept in algebra is that of a substructure that is a subset having the same structure as the superset. Definition 1.4.1. A subring of a ring R is a nonempty subset S that is also a ring under the same operations as R. If R is a field and S also a field then its a subfield. If S  R then S satisfies the same basic axioms, associativity and commutativity of addition for example. Therefore S will be a subring if it is nonempty and closed under the operations, that is closed under addition, multiplication and taking additive inverses. Lemma 1.4.2. A subset S of a ring R is a subring if and only if S is nonempty and whenever a; b 2 S we have a C b 2 S , a  b 2 S and ab 2 S.

Section 1.4 Subrings and Ideals

7

Example 1.4.3. Show that if n > 1 the set nZ is a subring of Z. Here clearly nZ is nonempty. Suppose a D nz1 ; b D nz2 are two element of nZ. Then a C b D nz1 C nz2 D n.z1 C z2 / 2 nZ a  b D nz1  nz2 D n.z1  z2 / 2 nZ ab D nz1  nz2 D n.nz1 z2 / 2 nZ: Therefore nZ is a subring. Example 1.4.4. Show that the set of real numbers of the form p S D ¹u C v 2 W u; v 2 Qº is a subring of p p p R. Here 1 C 2 2 S , so S is nonempty. Suppose a D u1 C v1 2, b D u2 C v2 2 are two element of S. Then p p p a C b D .u1 C v1 2/ C .u2 C v2 2/ D u1 C u2 C .v1 C v2 / 2 2 S p p p a  b D .u1 C v1 2/  .u2 C v2 2/ D u1  u2 C .v1  v2 / 2 2 S p p p a  b D .u1 C v1 2/  .u2 C v2 2/ D .u1 u2 C 2v1 v2 / C .u1 v2 C v1 u2 / 2 2 S: Therefore S is a subring. We will see this example later as an algebraic number field. In the following we are especially interested in special types of subrings called ideals. Definition 1.4.5. Let R be a ring and I  R. Then I is a (two-sided) ideal if the following properties holds: (1) I is nonempty. (2) If a; b 2 I then a ˙ b 2 I . (3) If a 2 I and r is any element of R then ra 2 I and ar 2 I . We denote the fact that I forms an ideal in R by I G R. Notice that if a; b 2 I , then from (3) we have ab 2 I and ba 2 I . Hence I forms a subring, that is each ideal is also a subring. ¹0º and the whole ring R are trivial ideals of R. If we assume that in (3) only ra 2 I then I is called a left ideal. Analogously we define a right ideal.

8

Chapter 1 Groups, Rings and Fields

Lemma 1.4.6. Let R be a commutative ring and a 2 R. Then the set hai D aR D ¹ar W r 2 Rº is an ideal of R. This ideal is called the principal ideal generated by a. Proof. We must verify the three properties of the definition. Since a 2 R we have that aR is nonempty. If u D ar1 ; v D ar2 are two elements of aR then u ˙ v D ar1 ˙ ar2 D a.r1 ˙ r2 / 2 aR so (2) is satisfied. Finally let u D ar1 2 aR and r 2 R. Then ru D rar1 D a.rr1 / 2 aR

and

ur D ar1 r D a.r1 r/ 2 aR:

Recall that a 2 hai if R has an identity. Notice that if n 2 Z then the principal ideal generated by n is precisely the ring nZ, that we have already examined. Hence for each n > 1 the subring nZ is actually an ideal. We can show more. Theorem 1.4.7. Any subring of Z is of the form nZ for some n. Hence each subring of Z is actually a principal ideal. Proof. Let S be a subring of Z. If S D ¹0º then S D 0Z so we may assume that S has nonzero elements. Since S is a subring if it has nonzero elements it must have positive elements (since it has the additive inverse of any element in it). Let S C be the set of positive elements in S . From the remarks above this is a nonempty set and so there must be a least positive element n. We claim that S D nZ. Let m be a positive element in S . By the division algorithm m D q n C r; where either r D 0 or 0 < r < n. Suppose that r ¤ 0. Then r D m  q n: Now m 2 S and n 2 S . Since S is a subring it is closed under addition so that q n 2 S . But S is a subring so m  q n 2 S. It follows that r 2 S. But this is a contradiction since n was the least positive element in S. Therefore r D 0 and m D q n. Hence each positive element in S is a multiple of n. Now let m be a negative element of S . Then m 2 S and m is positive. Hence m D q n and thus m D .q/n. Therefore every element of S is a multiple of n and so S D nZ. It follows that every subring of Z is of this form and therefore every subring of Z is an ideal.

Section 1.5 Factor Rings and Ring Homomorphisms

9

We mention that this is true in Z but not always true. For example Z is a subring of Q but not an ideal. An extension of the proof of Lemma 1.4.2 gives the following. We leave the proof as an exercise. Lemma 1.4.8. Let R be a commutative ring and a1 ; : : : ; an 2 R be a finite set of elements in R. Then the set ha1 ; : : : ; an i D ¹r1 a1 C r2 a2 C    C rn an W ri 2 Rº is an ideal of R. This ideal is called the ideal generated by a1 ; : : : ; an . Recall that a1 ; : : : ; an are in ha1 ; : : : ; an i if R has an identity. Theorem 1.4.9. Let R be a commutative ring with an identity 1 ¤ 0. Then R is a field if and only if the only ideals in R are ¹0º and R. Proof. Suppose that R is a field and I C R is an ideal. We must show that either I D ¹0º or I D R. Suppose that I ¤ ¹0º then we must show that I D R. Since I ¤ ¹0º there exists an element a 2 I with a ¤ 0. Since R is a field this element a has an inverse a1 . Since I is an ideal it follows that a1 a D 1 2 I . Let r 2 R then, since 1 2 I , we have r  1 D r 2 I . Hence R  I and hence R D I . Conversely suppose that R is a commutative ring with an identity whose only ideals are ¹0º and R. We must show that R is a field or equivalently that every nonzero element of R has a multiplicative inverse. Let a 2 R with a ¤ 0. Since R is a commutative ring and a ¤ 0, the principal ideal aR is a nontrivial ideal in R. Hence aR D R. Therefore the multiplicative identity 1 2 aR. It follows that there exists an r 2 R with ar D 1. Hence a has a multiplicative inverse and R must be a field.

1.5

Factor Rings and Ring Homomorphisms

Given an ideal I in a ring R we can build a new ring called the factor ring or quotient ring of R modulo I . The special condition on the subring I that rI  I and I r  I for all r 2 R, that makes it an ideal, is specifically to allow this construction to be a ring. Definition 1.5.1. Let I be an ideal in a ring R. Then a coset of I is a subset of R of the form r C I D ¹r C i W i 2 I º with r a fixed element of R.

10

Chapter 1 Groups, Rings and Fields

Lemma 1.5.2. Let I be an ideal in a ring R. Then the cosets of I partition R, that is any two cosets are either coincide or disjoint. We leave the proof to the exercises. Now on the set of all cosets of an ideal we will build a new ring. Theorem 1.5.3. Let I be an ideal in a ring R. Let R=I be the set of all cosets of I in R, that is R=I D ¹r C I W r 2 Rº: We define addition and multiplication on R=I in the following manner: .r1 C I / C .r2 C I / D .r1 C r2 / C I .r1 C I /  .r2 C I / D .r1  r2 / C I: Then R=I forms a ring called the factor ring of R modulo I . The zero element of R=I is 0 C I and the additive inverse of r C I is r C I . Further if R is commutative then R=I is commutative and if R has an identity then R=I has an identity 1 C I . Proof. The proofs that R=I satisfies the ring axioms under the definitions above is straightforward. For example .r1 C I / C .r2 C I / D .r1 C r2 / C I D .r2 C r1 / C I D .r2 C I / C .r1 C I / and so addition is commutative. What must be shown is that both addition and multiplication are well-defined. That is, if r1 C I D r10 C I and r2 C I D r20 C I then and

.r1 C I / C .r2 C I / D .r10 C I / C .r20 C I / .r1 C I /  .r2 C I / D .r10 C I /  .r20 C I /:

Now if r1 C I D r10 C I then r1 2 r10 C I and so r1 D r10 C i1 for some i1 2 I . Similarly if r2 C I D r20 C I then r2 2 r20 C I and so r2 D r20 C i2 for some i2 2 I . Then .r1 C I / C .r2 C I / D .r10 C i1 C I / C .r20 C i2 C I / D .r10 C I / C .r20 C I / since i1 C I D I and i2 C I D I . Similarly .r1 C I /  .r2 C I / D .r10 C i1 C I /  .r20 C i2 C I / D r10  r20 C r10 i2 C r20 i1 C r10 I C r20 I C I  I D .r10  r20 / C I since all the other products are in the ideal I .

Section 1.5 Factor Rings and Ring Homomorphisms

11

This shows that addition and multiplication are well-defined. It also shows why the ideal property is necessary. As an example let R be the integers Z. As we have seen each subring is an ideal and of the form nZ for some natural number n. The factor ring Z=nZ is called the residue class ring modulo n denoted Zn . Notice that we can take as cosets 0 C nZ; 1 C nZ; : : : ; .n  1/ C nZ: Addition and multiplication of cosets is then just addition and multiplication modulo n, as we can see, that this is just a formalization of the ring Zn , that we have already looked at. Recall that Zn is an integral domain if and only if n is prime and Zn is a field for precisely the same n. If n D 0 then Z=nZ is the same as Z. We now show that ideals and factor rings are closely related to certain mappings between rings. Definition 1.5.4. Let R and S be rings. Then a mapping f W R ! S is a ring homomorphism if f .r1 C r2 / D f .r1 / C f .r2 / for any r1 ; r2 2 R f .r1  r2 / D f .r1 /  f .r2 /

for any r1 ; r2 2 R:

In addition, (1) f is an epimorphism if it is surjective. (2) f is an monomorphism if it is injective. (3) f is an isomorphism if it is bijective, that is both surjective and injective. In this case R and S are said to be isomorphic rings which we denote by R Š S. (4) f is an endomorphism if R D S , that is a ring homomorphism from a ring to itself. (5) f is an automorphism if R D S and f is an isomorphism. Lemma 1.5.5. Let R and S be rings and let f W R ! S be a ring homomorphism. Then (1) f .0/ D 0 where the first 0 is the zero element of R and the second is the zero element of S. (2) f .r/ D f .r/ for any r 2 R. Proof. We obtain f .0/ D 0 from the equation f .0/ D f .0 C 0/ D f .0/ C f .0/. Hence 0 D f .0/ D f .r  r/ D f .r C .r// D f .r/ C f .r/, that is f .r/ D f .r/.

12

Chapter 1 Groups, Rings and Fields

Definition 1.5.6. Let R and S be rings and let f W R ! S be a ring homomorphism. Then the kernel of f is ker.f / D ¹r 2 R W f .r/ D 0º: The image of f , denoted im.f /, is the range of f within S. That is im.f / D ¹s 2 S W there exists r 2 R with f .r/ D sº: Theorem 1.5.7 (ring isomorphism theorem). Let R and S be rings and let f WR!S be a ring homomorphism. Then (1) ker.f / is an ideal in R, im.f / is a subring of S and R= ker.f / Š im.f /: (2) Conversely suppose that I is an ideal in a ring R. Then the map f W R ! R=I given by f .r/ D r C I for r 2 R is a ring homomorphism whose kernel is I and whose image is R=I . The theorem says that the concepts of ideal of a ring and kernel of a ring homomorphism coincide, that is each ideal is the kernel of a homomorphism and the kernel of each ring homomorphism is an ideal. Proof. Let f W R ! S be a ring homomorphism and let I D ker.f /. We show first that I is an ideal. If r1 ; r2 2 I then f .r1 / D f .r2 / D 0. It follows from the homomorphism property that f .r1 ˙ r2 / D f .r1 / ˙ f .r2 / D 0 C 0 D 0 f .r1  r2 / D f .r1 /  f .r2 / D 0  0 D 0: Therefore I is a subring. Now let i 2 I and r 2 R. Then f .r  i / D f .r/  f .i / D f .r/  0 D 0

and

f .i  r/ D f .i/  f .r/ D 0  f .r/ D 0

and hence I is an ideal. Consider the factor ring R=I . Let f  W R=I ! im.f / by f  .r C I / D f .r/. We show that f  is an isomorphism. First we show that it is well-defined. Suppose that r1 C I D r2 C I then r1  r2 2 I D ker.f /. It follows that f .r1  r2 / D 0 so f .r1 / D f .r2 /. Hence f  .r1 C I / D f  .r2 C I / and the map f  is well-defined.

13

Section 1.6 Fields of Fractions

Now f  ..r1 C I / C .r2 C I // D f  ..r1 C r2 / C I / D f .r1 C r2 / D f .r1 / C f .r2 / D f  .r1 C I / C f  .r2 C I / and f  ..r1 C I /  .r2 C I // D f  ..r1  r2 / C I / D f .r1  r2 / D f .r1 /  f .r2 / D f  .r1 C I /  f  .r2 C I /: Hence f  is a homomorphism. We must now show that it is injective and surjective. Suppose that f  .r1 CI / D f  .r2 CI /. Then f .r1 / D f .r2 / so that f .r1 r2 / D 0. Hence r1  r2 2 ker.f / D I . Therefore r1 2 r2 C I and thus r1 C I D r2 C I and the map f  is injective. Finally let s 2 im.f /. Then there exists and r 2 R such that f .r/ D s. Then f  .r C I / D s and the map f  is surjective and hence an isomorphism. This proves the first part of the theorem. To prove the second part let I be an ideal in R and R=I the factor ring. Consider the map f W R ! R=I given by f .r/ D r C I . From the definition of addition and multiplication in the factor ring R=I it is clear that this is a homomorphism. Consider the kernel of f . If r 2 ker.f / then f .r/ D r C I D 0 D 0 C I . This implies that r 2 I and hence the kernel of this map is exactly the ideal I completing the theorem. Theorem 1.5.7 is called the ring isomorphism theorem or the first ring isomorphism theorem. We mention that there is an analogous theorem for each algebraic structure. In particular for groups and vector spaces. We will mention the result for groups in Section 1.8.

1.6

Fields of Fractions

The integers are an integral domain and the rationals Q are a field that contains the integers. First we show that Q is the smallest field containing Z. Theorem 1.6.1. The rationals Q are the smallest field containing the integers Z. That is if Z  F  Q with F a subfield of Q then F D Q. Proof. Since Z  F we have m; n 2 F for any two integers m; n. Since F is a subfield, it is closed under taking division, that is taking multiplicative inverses and hence the fraction m n 2 F . Since each element of Q is such a fraction it follows that Q  F . Since F  Q it follows that F D Q.

14

Chapter 1 Groups, Rings and Fields

Notice that to construct the rationals from the integers we form all the fractions m2 1 with n ¤ 0 and where m n1 D n2 if m1 n2 D n1 m2 . We then do the standard operations on fractions. If we start with any integral domain D we can mimic this construction to build a field of fractions from D that is the smallest field containing D. m n

Theorem 1.6.2. Let D be an integral domain. Then there is a field F containing D, called the field of fractions for D, such that each element of F is a fraction from D, that is an element of the form d1 d21 with d1 ; d2 2 D. Further F is unique up to isomorphism and is the smallest field containing D. Proof. The proof is just the mimicking of the construction of the rationals from the integers. Let F ? D ¹.d1 ; d2 / W d1 ; d2 ¤ 0; d1 ; d2 2 Dº: Define on F ? the equivalence relation .d1 ; d2 / D .d10 ; d20 / if d1 d20 D d2 d10 : Let F be the set of equivalence classes and define addition and multiplication in the usual manner as for fractions where the result is the equivalence class. .d1 ; d2 / C .d3 ; d4 / D .d1 d4 C d2 d3 ; d2 d4 / .d1 ; d2 /  .d3 ; d4 / D .d1 d3 ; d2 d4 /: It is now straightforward to verify the ring axioms for F . The inverse of .d1 ; 1/ is .1; d1 / for d1 ¤ 0 in D. As with Z we identify the elements of F as fractions dd12 . The proof that F is the smallest field containing D is the same as for Q from Z. As examples we have that Q is the field of fractions for Z. A familiar but less common example is the following. Let RŒx be the set of polynomials over the real numbers R. It can be shown that RŒx forms an integral domain. The field of fractions consists of all formal functions f .x/ where f .x/; g.x/ are real polynomials with g.x/ ¤ 0. The corresponding field g.x/ of fractions is called the field of rational functions over R and is denoted R.x/.

1.7

Characteristic and Prime Rings

We saw in the last section that Q is the smallest field containing the integers. Since any subfield of Q must contain the identity, it follows that any nontrivial subfield of Q must contain the integers and hence be all of Q. Therefore Q has no nontrivial subfields. We say that Q is a prime field.

Section 1.7 Characteristic and Prime Rings

15

Definition 1.7.1. A field F is a prime field if F contains no nontrivial subfields. Lemma 1.7.2. Let K be any field. Then K contains a prime field F as a subfield. Proof. Let K1 ; K2 be subfields of K. If k1 ; k2 2 K1 \K2 then k1 ˙k2 2 K1 since K1 is a subfield and k1 ˙ k2 2 K2 since K2 is a subfield. Therefore k1 ˙ k2 2 K1 \ K2 . Similarly k1 k21 2 K1 \ K2 . It follows that K1 \ K2 is again a subfield. Now let F be the intersection of all subfields of K. From the argument above F is a subfield and the only nontrivial subfield of F is itself. Hence F is a prime field. Definition 1.7.3. Let R be a commutative ring with an identity 1 ¤ 0. The smallest positive integer n such that n  1 D 1 C 1 C    C 1 D 0 is called the characteristic of R. If there is no such n, then R has characteristic 0. We denote the characteristic by char.R/. Notice first that the characteristic of Z; Q; R are all zero. Further the characteristic of Zn is n. Theorem 1.7.4. Let R be an integral domain. Then the characteristic of R is either 0 or a prime. In particular the characteristic of a field is zero or a prime. Proof. Suppose that R is an integral domain and char.R/ D n ¤ 0. Suppose that n D mk with 1 < m < n, 1 < k < n. Then n  1 D 0 D .m  1/.k  1/. Since R is an integral domain we have no zero divisors and hence m  1 D 0 or k  1 D 0. However this is a contradiction since n is the least positive integer such that n  1 D 0. Therefore n must be a prime. We have seen that every field contains a prime field. We extend this. Definition 1.7.5. A commutative ring R with an identity 1 ¤ 0 is a prime ring if the only subring containing the identity is the whole ring. Clearly both the integers Z and the modular integers Zn are prime rings. In fact up to isomorphism they are the only prime rings. Theorem 1.7.6. Let R be a prime ring. If char.R/ D 0 then R Š Z, while if char.R/ D n > 0 then R Š Zn . Proof. Suppose that char.R/ D 0. Let S D ¹r D m  1 W r 2 R; m 2 Zº. Then S is a subring of R containing the identity (see the exercises) and hence S D R. However the map m  1 ! m gives an isomorphism from S to Z. It follows that R is isomorphic to Z. If char.R/ D n > 0 the proof is identical. Since n  1 D 0 the subring S of R defined above is all of R and isomorphic to Zn .

16

Chapter 1 Groups, Rings and Fields

Theorem 1.7.6 can be extended to fields with Q taking the place of Z and Zp , with p a prime, taking the place of Zn . Theorem 1.7.7. Let K be a prime field. If K has characteristic 0 then K Š Q while if K has characteristic p then K Š Zp . Proof. The proof is identical to that of Theorem 1.7.6; however we consider the smallest subfield K1 of K containing S. We mention that there can be infinite fields of characteristic p. Consider for example the field of fractions of the polynomial ring Zp Œx. This is the field of rational functions with coefficients in Zp . We give a theorem on fields of characteristic p that will be important much later when we look at Galois theory. Theorem 1.7.8. Let K be a field of characteristic p. Then the mapping  W K ! K given by .k/ D k p is an injective endomorphism of K. In particular .a C b/p D ap C b p for any a; b 2 K. This mapping is called the Frobenius homomorphism of K. Further if K is finite,  is an automorphism. Proof. We first show that  is a homomorphism. Now .ab/ D .ab/p D ap b p D .a/.b/: We need a little more work for addition. ! ! p p1 X X p p i pi p p ab ai b pi C b p .a C b/ D .a C b/ D Da C i i i D0

i D1

by the binomial expansion which holds in any commutative ring. However ! p p.p1 /    .p  i C 1/ D i i  .i  1/    1   p  and it is clear that pj i for 1  i  p  1. Hence in K we have pi  1 D 0 and so we have .a C b/ D .a C b/p D ap C b p D .a/ C .b/: Therefore  is a homomorphism. Further  is always injective. To see this suppose that .x/ D .y/. Then .x  y/ D 0 H) .x  y/p D 0: But K is a field so there are no zero divisors so we must have x  y D 0 or x D y. If K is finite and  is injective it must also be surjective and hence an automorphism of K.

17

Section 1.8 Groups

1.8

Groups

We close this first chapter by introducing some basic definitions and results from group theory, that mirror the results, that were presented for rings and fields. We will look at group theory in more detail later in the book. Proofs will be given at that point. Definition 1.8.1. A group G is a set with one binary operation (which we will denote by multiplication) such that (1) The operation is associative. (2) There exists an identity for this operation. (3) Each g 2 G has an inverse for this operation. If, in addition, the operation is commutative, the group G is called an abelian group. The order of G is the number of elements in G, denoted by jGj. If jGj < 1; G is a finite group otherwise G is an infinite group. Groups most often arise from invertible mappings of a set onto itself. Such mappings are called permutations. Theorem 1.8.2. The group of all permutations on a set A forms a group called the symmetric group on A which we denote by SA . If A has more than 2 elements then SA is nonabelian. Definition 1.8.3. Let G1 and G2 be groups. Then a mapping f W G1 ! G2 is a (group) homomorphism if f .g1 g2 / D f .g1 /f .g2 /

for any g1 ; g2 2 G1 :

As with rings we have further (1) f is an epimorphism if it is surjective. (2) f is an monomorphism if it is injective. (3) f is an isomorphism if it is bijective, that is both surjective and injective. In this case G1 and G2 are said to be isomorphic groups, which we denote by G1 Š G2 . (4) f is an endomorphism if G1 D G2 , that is a homomorphism from a group to itself. (5) f is an automorphism if G1 D G2 and f is an isomorphism.

18

Chapter 1 Groups, Rings and Fields

Lemma 1.8.4. Let G1 and G2 be groups and let f W G1 ! G2 be a homomorphism. Then (a) f .1/ D 1 where the first 1 is the identity element of G1 and the second is the identity element of G2 . (b) f .g 1 / D .f .g//1 for any g 2 G1 . If A is a set, jAj denotes the size of A. Theorem 1.8.5. If A1 and A2 are sets with jA1 j D jA2 j then SA1 Š SA2 . If jAj D n with n finite we call SA the symmetric group on n elements which we denote by Sn . Further we have jSn j D nŠ. Subgroups are defined in an analogous manner to subrings. Special types of subgroups called normal subgroups take the place in group theory that ideals play in ring theory. Definition 1.8.6. A subset H of a group G is a subgroup if H ¤ ; and H forms a group under the same operation as G. Equivalently, H is a subgroup if H ¤ ; and H is closed under the operation and inverses. Definition 1.8.7. If H is a subgroup of a group G, then a left coset of H is a subset of G of the form gH D ¹gh W h 2 H º. A right coset of H is a subset of G of the form Hg D ¹hg W h 2 H º. As with rings the cosets of a subgroup partition a group. We call the number of right cosets of a subgroup H in a group G then index of H in G, denoted jG W H j. One can prove that the number of right cosets is equal to the number of left cosets. For finite groups we have the following beautiful result called Lagrange’s theorem. Theorem 1.8.8 (Lagrange’s theorem). Let G be a finite group and H a subgroup. Then the order of H divides the order of G. In particular jGj D jH jjG W H j: Normal subgroups take the place of ideals in group theory. Definition 1.8.9. A subgroup H of a group G is a normal subgroup, denoted H C G, if every left coset of H is also a right coset, that is gH D Hg for each g 2 G. Note that this does not say that g and H commute elementwise, just that the subsets gH and Hg are the same. Equivalently H is normal if g1 Hg D H for any g 2 G. Normal subgroups allow us to construct factor groups just as ideals allowed us to construct factor rings.

19

Section 1.9 Exercises

Theorem 1.8.10. Let H be a normal subgroup of a group G. Let G=H be the set of all cosets of H in G, that is G=H D ¹gH W g 2 Gº: We define multiplication on G=H in the following manner .g1 H /.g2 H / D g1 g2 H: Then G=H forms a group called the factor group or quotient group of G modulo H . The identity element of G=H is 1H and the inverse of gH is g 1 H . Further if G is abelian then G=H is also abelian. Finally as with rings normal subgroups, factor groups are closely tied to homomorphisms. Definition 1.8.11. Let G1 and G2 be groups and let f W G1 ! G2 be a homomorphism. Then the kernel of f , denoted ker.f /, is ker.f / D ¹g 2 G1 W f .g/ D 1º: The image of f , denoted im.f /, is the range of f within G2 . That is im.f / D ¹h 2 G2 W there exists g 2 G1 with f .g/ D hº: Theorem 1.8.12 (group isomorphism theorem). Let G1 and G2 be groups and let f W G1 ! G2 be a homomorphism. Then (1) ker.f / is a normal subgroup in G1 . im.f / is a subgroup of G2 and G1 = ker.f / Š im.f /: (2) Conversely suppose that H is a normal subgroup of a group G. Then the map f W G ! G=H given by f .g/ D gH for g 2 G is a homomorphism whose kernel is H and whose image is G=H .

1.9

Exercises

1. Let  W K ! R be a homomorphism from a field K to a ring R. Show: Either .a/ D 0 for all a 2 K or  is a monomorphism. 2. Let R be a ring and M ¤ ; an arbitrary set. Show that the following are equivalent: (i) The ring of all mappings from M to R is a field. (ii) M contains only one element and R is a field.

20

Chapter 1 Groups, Rings and Fields

3. Let  be a set of prime numbers. Define ² ³ a Q D W all prime divisors of b are in  : b (i) Show that Q is a subring of Q. (ii) Let R be a subring of Q and let 1 2 R. b

a b

2 R with coprime integers a; b. Show that

(iii) Determine all subrings R of Q. (Hint: Consider the set of all prime divisors of denominators of reduced elements of R.) 4. Prove Lemma 1.5.2. 5. Let R be a commutative ring with an identity 1 2 R. Let A, B and C be ideals in R. A C B WD ¹a C b W a 2 A; b 2 Bº and AB WD .¹ab W a 2 A; b 2 Bº/. Show: (i) A C B G R, A C B D .A [ B/ (ii) AB D ¹a1 b1 C    C an bn W n 2 N; ai 2 A; bi 2 Bº, AB  A \ B (iii) A.B C C/ D AB C AC , .A C B/C D AB C BC , .AB/C D A.BC / (iv) A D R , A \ R ¤ ; (v) a; b 2 R ) hai C hbi D ¹xa C yb W x; y 2 Rº (vi) a; b 2 R ) haihbi D habi. Here hai D Ra D ¹xa W x 2 Rº. 6. Solve the following congruence: 3x  5 mod 7: Is this congruence also solvable mod 17? 7. Show that the set of 2  2 matrices over a ring R forms a ring. 8. Prove Lemma 1.4.8. 9. Prove that if R is a ring with identity and S D ¹r D m  1 W r 2 R; m 2 Zº then S is a subring of R containing the identity.

Chapter 2

Maximal and Prime Ideals

2.1

Maximal and Prime Ideals

In the first chapter we defined ideals I in a ring R and then the factor ring R=I of R modulo the ideal I . We saw further that if R is commutative then R=I is also commutative and if R has an identity then so does R=I . This raises further questions concerning the structure of factor rings. In particular we can ask under what conditions does R=I form an integral domain and under what conditions does R=I form a field. These questions lead us to define certain special properties of ideals, called prime ideals and maximal ideals. For motivation let us look back at the integers Z. Recall that each proper ideal in Z has the form nZ for some n > 1 and the resulting factor ring Z=nZ is isomorphic to Zn . We proved the following result. Theorem 2.1.1. Zn D Z=nZ is an integral domain if and only if n D p a prime. Further Zn is a field again if and only if n D p is a prime. Hence for the integers Z a factor ring is a field if and only if it is an integral domain. We will see later that this is not true in general. However what is clear is that the special ideals nZ leading to integral domains and fields are precisely when n is a prime. We look at the ideals pZ with p a prime in two different ways and then use these in subsequent sections to give the general definitions. We first need a famous result, Euclid’s lemma, from number theory. For integers a; b the notation ajb means that a divides b. Lemma 2.1.2 (Euclid). If p is a prime and pjab then pja or pjb. Proof. Recall that the greatest common divisor or GCD of two integers a; b is an integer d > 0 such that d is a common divisor of both a and b and if d1 is another common divisor of a and b then d1 jd . We express the GCD of a; b by d D .a; b/. It is known that for any two integers a; b their GCD exists and is unique and further is the least positive linear combination of a and b, that is the least positive integer of the form ax C by for integers x; y. The integers a; b are relatively prime if their GCD is 1, .a; b/ D 1. In this case 1 is a linear combination of a and b. Now suppose pjab where p is a prime. If p does not divide a then since the only positive divisors of p are 1 and p it follows that .a; p/ D 1. Hence 1 is expressible

22

Chapter 2 Maximal and Prime Ideals

as a linear combination of a and p. That is ax C py D 1 for some integers x; y. Multiply through by b, so that abx C pby D b: Now pjab so pjabx and pjpby. Therefore pjabx C pby, that is, pjb. We now recast this lemma in two different ways in terms of the ideal pZ. Notice that pZ consists precisely of all the multiples of p. Hence pjab is equivalent to ab 2 pZ. Lemma 2.1.3. If p is a prime and ab 2 pZ then a 2 pZ or b 2 pZ. This conclusion will be taken as the definition of a prime ideal in the next section. Lemma 2.1.4. If p is a prime and pZ  nZ then n D 1 or n D p. That is, every ideal in Z containing pZ with p a prime is either all of Z or pZ. Proof. Suppose that pZ  nZ. Then p 2 nZ so p is a multiple of n. Since p is a prime it follows easily that either n D 1 or n D p. In Section 2.3 the conclusion of this lemma will be taken as the definition of a maximal ideal.

2.2

Prime Ideals and Integral Domains

Motivated by Lemma 2.1.3 we make the following general definition for commutative rings R with identity. Definition 2.2.1. Let R be a commutative ring. An ideal P in R with P ¤ R is a prime ideal if whenever ab 2 P with a; b 2 R then either a 2 P or b 2 P . This property of an ideal is precisely what is necessary and sufficient to make the factor ring R=I an integral domain. Theorem 2.2.2. Let R be a commutative ring with an identity 1 ¤ 0 and let P be a nontrivial ideal in R. Then P is a prime ideal if and only if the factor ring R=P is an integral domain. Proof. Let R be a commutative ring with an identity 1 ¤ 0 and let P be a prime ideal. We show that R=P is an integral domain. From the results in the last chapter we have that R=P is again a commutative ring with an identity. Therefore we must show that there are no zero divisors in R=P . Suppose that .a C I /.b C I / D 0 in R=P . The zero element in R=P is 0 C P and hence .a C P /.b C P / D 0 D 0 C P H) ab C P D 0 C P H) ab 2 P:

Section 2.2 Prime Ideals and Integral Domains

23

However P is a prime ideal so we must then have a 2 P or b 2 P . If a 2 P then a C P D P D 0 C P so a C P D 0 in R=P . The identical argument works if b 2 P . Therefore there are no zero divisors in R=P and hence R=P is an integral domain. Conversely suppose that R=P is an integral domain. We must show that P is a prime ideal. Suppose that ab 2 P . Then .a C P /.b C P / D ab C P D 0 C P . Hence in R=P we have .a C P /.b C P / D 0: However R=P is an integral domain so it has no zero divisors. It follows that either a C P D 0 and hence a 2 P or b C P D 0 and b 2 P . Therefore either a 2 P or b 2 P so P is a prime ideal. In a commutative ring R we can define a multiplication of ideals. We then obtain an exact analog of Euclid’s lemma. Since R is commutative each ideal is 2-sided. Definition 2.2.3. Let R be a commutative ring with an identity 1 ¤ 0 and let A and B be ideals in R. Define AB D ¹a1 b1 C    C an bn W ai 2 A; bi 2 B; n 2 Nº: That is AB is the set of finite sums of products ab with a 2 A and b 2 B. Lemma 2.2.4. Let R be a commutative ring with an identity 1 ¤ 0 and let A and B be ideals in R. Then AB is an ideal. Proof. We must verify that AB is a subring and that it is closed under multiplication from R. Le r1 ; r2 2 AB. Then r1 D a1 b1 C    C an bn and Then

0 0 bm r2 D a10 b10 C    C am

for some ai 2 A; bi 2 B for some ai0 2 A; bi0 2 B:

0 0 r1 ˙ r2 D a1 b1 C    C an bn ˙ a10 b10 ˙    ˙ am bm

which is clearly in AB. Further 0 0 bm : r1  r2 D a1 b1 a10 b10 C    C an bn am

Consider for example the first term a1 b1 a10 b10 . Since R is commutative this is equal to .a1 a10 /.b1 b10 /: Now a1 a10 2 A since A is a subring and b1 b10 2 B since B is a subring. Hence this term is in AB. Similarly for each of the other terms. Therefore r1 r2 2 AB and hence AB is a subring.

24

Chapter 2 Maximal and Prime Ideals

Now let r 2 R and consider rr1 . This is then rr1 D ra1 b1 C    C ran bn : Now rai 2 A for each i since A is an ideal. Hence each summand is in AB and then rr1 2 AB. Therefore AB is an ideal. Lemma 2.2.5. Let R be a commutative ring with an identity 1 ¤ 0 and let A and B be ideals in R. If P is a prime ideal in R then AB  P implies that A  P or B  P. Proof. Suppose that AB  P with P a prime ideal and suppose that B is not contained in P . We show that A  P . Since AB  P each product ai bj 2 P . Choose a b 2 B with b … P and let a be an arbitrary element of A. Then ab 2 P . Since P is a prime ideal this implies either a 2 P or b 2 P . But by assumption b … P so a 2 P . Since a was arbitrary we have A  P .

2.3

Maximal Ideals and Fields

Now, motivated by Lemma 2.1.4 we define a maximal ideal. Definition 2.3.1. Let R be a ring and I an ideal in R. Then I is a maximal ideal if I ¤ R and if J is an ideal in R with I  J then I D J or J D R. If R is a commutative ring with an identity this property of an ideal I is precisely what is necessary and sufficient so that R=I is a field. Theorem 2.3.2. Let R be a commutative ring with an identity 1 ¤ 0 and let I be an ideal in R. Then I is a maximal ideal if and only if the factor ring R=I is a field. Proof. Suppose that R is a commutative ring with an identity 1 ¤ 0 and let I be an ideal in R. Suppose first that I is a maximal ideal and we show that the factor ring R=I is a field. Since R is a commutative ring with an identity the factor ring R=I is also a commutative ring with an identity. We must show then that each nonzero element of R=I has a multiplicative inverse. Suppose then that r D r CI 2 R=I is a nonzero element of R=I . It follows that r … I . Consider the set hr; I i D ¹rx C i W x 2 R; i 2 I º. This is also an ideal (see exercises) called the ideal generated by r and I , denoted hr; I i. Clearly I  hr; I i and since r … I and r D r  1 C 0 2 hr; I i it follows that hr; I i ¤ I . Since I is a maximal ideal it follows that hr; I i D R the whole ring. Hence the identity element 1 2 hr; I i and so there exist elements x 2 R and i 2 I such that 1 D rx C i . But then 1 2 .r C I /.x C I / and so 1 C I D .r C I /.x C I /.

Section 2.4 The Existence of Maximal Ideals

25

Since 1 C I is the multiplicative identity of R=I is follows that x C I is the multiplicative inverse of r C I in R=I . Since r C I was an arbitrary nonzero element of R=I it follows that R=I is a field. Now suppose that R=I is a field for an ideal I . We show that I must be maximal. Suppose then that I1 is an ideal with I  I1 and I ¤ I1 . We must show that I1 is all of R. Since I ¤ I1 there exists an r 2 I1 with r … I . Therefore the element r C I is nonzero in the factor ring R=I and since R=I is a field it must have a multiplicative inverse x C I . Hence .r C I /.x C I / D rx C I D 1 C I and therefore there is an i 2 I with 1 D rx C i. Since r 2 I1 and I1 is an ideal we get that rx 2 I1 . Further since I  I1 it follows that rx C i 2 I1 and so 1 2 I1 . If r1 is an arbitrary element of R then r1  1 D r1 2 I1 . Hence R  I1 and so R D I1 . Therefore I is a maximal ideal. Recall that a field is already an integral domain. Combining this with the ideas of prime and maximal ideals we obtain: Theorem 2.3.3. Let R be a commutative ring with an identity 1 ¤ 0. Then each maximal ideal is a prime ideal Proof. Suppose that R is a commutative ring with an identity and I is a maximal ideal in R. Then from Theorem 2.3.2 we have that the factor ring R=I is a field. But a field is an integral domain so R=I is an integral domain. Therefore from Theorem 2.2.2 we have that I must be a prime ideal. The converse is not true in general. That is there are prime ideals that are not maximal. Consider for example R D Z the integers and I D ¹0º. Then I is an ideal and R=I D Z=¹0º Š Z is an integral domain. Hence ¹0º is a prime ideal. However Z is not a field so ¹0º is not maximal. Note however that in the integers Z a proper ideal is maximal if and only if it is a prime ideal.

2.4

The Existence of Maximal Ideals

In this section we prove that in any ring R with an identity there do exist maximal ideals. Further given an ideal I ¤ R then there exists a maximal ideal I0 such that I  I0 . To prove this we need three important equivalent results from logic and set theory. First recall that a partial order  on a set S is a reflexive, transitive relation on S. That is a  a for all a 2 S and if a  b; b  c then a  c. This is a “partial” order since there may exist elements a 2 S where neither a  b nor b  a. If A is any set then it is clear that containment of subsets is a partial order on the power set P .A/. If  is a partial order on a set M , then a chain on M is a subset K  M such that a; b 2 K implies that a  b or b  a. A chain on M is bounded if there exists an

26

Chapter 2 Maximal and Prime Ideals

m 2 M such that k  m for all k 2 K. The element m is called an upper bound for K. An element m0 2 M is maximal if whenever m 2 M with m0  m then m D m0 . We now state the three important results from logic. Zorn’s lemma. If each chain of M has an upper bound in M then there is at least one maximal element in M . Axiom of well-ordering. Each set M can be well-ordered, such that each nonempty subset of M contains a least element. i 2 I º be a nonempty collection of nonempty sets. Then Axiom of choice. Let ¹Mi W S there is a mapping f W I ! i 2I Mi with f .i/ 2 Mi for all i 2 I . The following can be proved. Theorem 2.4.1. Zorn’s lemma, the axiom of well-ordering and the axiom of choice are all equivalent. We now show the existence of maximal ideals in commutative rings with identity. Theorem 2.4.2. Let R be a commutative ring with an identity 1 ¤ 0 and let I be an ideal in R with I ¤ R. Then there exists a maximal ideal I0 in R with I  I0 . In particular a ring with an identity contains maximal ideals. Proof. Let I be an ideal in the commutative ring R. We must show that there exists a maximal ideal I0 in R with I  I0 . Let M D ¹X W X is an ideal with I  X ¤ Rº: Then M is partially ordered by containment. We want to show first that each chain in M has a maximal element. If K D ¹Xj W Xj 2 M; j 2 J º is a chain let [ X0 D Xj : j 2J

If a; b 2 X 0 then there exists an i; j 2 J with a 2 Xi ; b 2 Xj . Since K is a chain either Xi  Xj or Xj  Xi . Without loss of generality suppose that Xi  Xj so that a; b 2 Xj . Then a ˙ b 2 Xj  X 0 and ab 2 Xj  X 0 since Xj is an ideal. Further if r 2 R then ra 2 Xj  X 0 since Xj is an ideal. Therefore X 0 is an ideal in R. Since Xj ¤ R it follows that 1 … Xj for all j 2 J . Therefore 1 … X 0 and so 0 ¤ R. It follows that under the partial order of containment X 0 is an upper bound X for K. We now use Zorn’s lemma. From the argument above we have that each chain has a maximal element. Hence for an ideal I the set M above has a maximal element. This maximal element I0 is then a maximal ideal containing I .

Section 2.5 Principal Ideals and Principal Ideal Domains

2.5

27

Principal Ideals and Principal Ideal Domains

Recall again that in the integers Z each ideal I is of the form nZ for some integer n. Hence in Z each ideal can be generated by a single element. Lemma 2.5.1. Let R be a commutative ring and a1 ; : : : ; an be elements of R. Then the set ha1 ; : : : ; an i D ¹r1 a1 C    C rn an W ri 2 Rº forms an ideal in R called the ideal generated by a1 ; : : : ; an . Proof. The proof is straightforward. Let a D r1 a1 C    C rn an ;

b D s1 a1 C    C sn an

with r1 ; : : : ; rn ; s1 ; : : : ; sn elements of R, be two elements of ha1 ; : : : ; an i. Then a ˙ b D .r1 ˙ s1 /a1 C    C .rn ˙ sn /an 2 ha1 ; : : : ; an i ab D .r1 s1 a1 /a1 C .r1 s2 a1 /a2 C    C .rn sn an /an 2 ha1 ; : : : ; an i so ha1 ; : : : ; an i forms a subring. Further if r 2 R we have ra D .rr1 /a1 C    C .rrn /an 2 ha1 ; : : : ; an i and so ha1 ; : : : ; an i is an ideal. Definition 2.5.2. Let R be a commutative ring. An ideal I  R is a principal ideal if it has a single generator. That is I D hai D aR

for some a 2 R:

We now restate Theorem 1.4.7 of Chapter 1. Theorem 2.5.3. Every nonzero ideal in Z is a principal ideal. Proof. Every ideal I in Z is of the form nZ. This is the principal ideal generated by n. Definition 2.5.4. A principal ideal domain or PID is an integral domain in which every ideal is principal. Corollary 2.5.5. The integers Z are a principal ideal domain. We mention that the set of polynomials KŒx with coefficients from a field K is also a principal ideal domain. We will return to this in the next chapter. Not every integral domain is a PID. Consider KŒx; y the set of polynomials over K in two variables x; y. Let I consist of all the polynomials with zero constant term.

28

Chapter 2 Maximal and Prime Ideals

Lemma 2.5.6. The set I in KŒx; y as defined above is an ideal but not a principal ideal. Proof. We leave the proof that I forms an ideal to the exercises. To show that it is not a principal ideal suppose I D hp.x; y/i. Now the polynomial q.x/ D x has zero constant term so q.x/ 2 I . Hence p.x; y/ cannot be a constant polynomial. Further if p.x; y/ had any terms with y in them there would be no way to multiply p.x; y/ by a polynomial h.x; y/ and obtain just x. Therefore p.x; y/ can contain no terms with y in them. But the same argument using s.y/ D y shows that p.x; y/ cannot have any terms with x in them. Therefore there can be no such p.x; y/ generating I and so I is not principal and KŒx; y is not a principal ideal domain.

2.6

Exercises

1. Consider the set hr; I i D ¹rx C i W x 2 R; i 2 I º where I is an ideal. Prove that this is also an ideal called the ideal generated by r and I , denoted hr; I i. 2. Let R and S be commutative rings and let  W R ! S be a ring epimorphism. Let M be a maximal ideal in R. Show: .M/ is a maximal ideal in S if and only if ker./  M. Is .M/ always a prime ideal of S? 3. Let A1 ; : : : ; A t be ideals of a commutative ring R. Let P be a prime ideal of R. Show: T (i) tiD1 Ai  P ) Aj  P for at least one index j . T (ii) tiD1 Ai D P ) Aj D P for at least one index j . 4. Which of the following ideals A are prime ideals of R? Which are maximal ideals? (i) A D .x/, R D ZŒx. (ii) A D .x 2 /, R D ZŒx. p p (iii) A D .1 C 5/, R D ZŒ 5. (iv) A D .x; y/, R D QŒx; y. p 5. Let w D 12 .1 C 3/. Show that h2i is a prime ideal and even a maximal ideal of ZŒw, but h2i is neither a prime ideal nor a maximal ideal of ZŒi. 6. Let R D ¹ ab W a; b 2 Z; b oddº. Show that R is a subring of Q and that there is only one maximal ideal M in R. 7. Let R be a ring with an identity. Let x; y 2 R and x ¤ 0 not be a zero divisor. Further let hxi be a prime ideal with hxi  hyi ¤ R. Show that hxi D hyi. 8. Consider KŒx; y the set of polynomials over K in two variables x; y. Let I consist of all the polynomials with zero constant term. Prove that the set I is an ideal.

Chapter 3

Prime Elements and Unique Factorization Domains

3.1

The Fundamental Theorem of Arithmetic

The integers Z have served as much of our motivation for properties of integral domains. In the last chapter we saw that Z is a principal ideal domain and furthermore that prime ideals ¤ ¹0º are maximal. From the viewpoint of the multiplicative structure of Z and the viewpoint of classical number theory the most important property of Z is the fundamental theorem of arithmetic. This states that any integer n ¤ 0 is uniquely expressible as a product of primes where uniqueness is up to ordering and the introduction of ˙1, that is units. In this chapter we show that this property is not unique to the integers and there are many other integral domains where this also holds. These are called unique factorization domains and we will present several examples. First we review the fundamental theorem of arithmetic, its proof and several other ideas from classical number theory. Theorem 3.1.1 (fundamental theorem of arithmetic). Given any integer n ¤ 0 there is a factorization n D cp1 p2    pk where c D ˙1 and p1 ; : : : ; pn are primes. Further this factorization is unique up to the ordering of the factors. There are two main ingredients that go into the proof; induction and Euclid’s lemma. We presented this in the last chapter. In turn however Euclid’s lemma depends upon the existence of greatest common divisors and their linear expressibility. Therefore to begin we present several basic ideas from number theory. The starting point for the theory of numbers is divisibility. Definition 3.1.2. If a; b are integers we say that a divides b, or that a is a factor or divisor of b, if there exists an integer q such that b D aq. We denote this by ajb. b is then a multiple of a. If b > 1 is an integer whose only factors are ˙1; ˙b then b is a prime, otherwise b > 1 is composite. The following properties of divisibility are straightforward consequences of the definition.

30

Chapter 3 Prime Elements and Unique Factorization Domains

Lemma 3.1.3. The following properties hold: (1) ajb ) ajbc for any integer c. (2) ajb and bjc implies ajc. (3) ajb and ajc implies that aj.bx C cy/ for any integers x; y. (4) ajb and bja implies that a D ˙b. (5) If ajb and a > 0; b > 0 then a  b. (6) ajb if and only if cajcb for any integer c ¤ 0. (7) aj0 for all a 2 Z and 0ja only for a D 0. (8) aj ˙ 1 only for a D ˙1. (9) a1 jb1 and a2 jb2 implies that a1 a2 jb1 b2 . If b; c; x; y are integers then an integer bx C cy is called a linear combination of b; c. Thus part (3) of Lemma 3.1.3 says that if a is a common divisor of b; c then a divides any linear combination of b and c. Further, note that if b > 1 is a composite then there exists x > 0 and y > 0 such that b D xy and from part (5) we must have 1 < x < b, 1 < y < b. In ordinary arithmetic, given a; b we can always attempt to divide a into b. The next result called the division algorithm says that if a > 0 either a will divide b or the remainder of the division of b by a will be less than a. Theorem 3.1.4 (division algorithm). Given integers a; b with a > 0 then there exist unique integers q and r such that b D qa C r where either r D 0 or 0 < r < a. One may think of q and r as the quotient and remainder respectively when dividing b by a. Proof. Given a; b with a > 0 consider the set S D ¹b  qa  0 W q 2 Zº: If b > 0 then b C a  0 and the sum is in S. If b  0 then there exists a q > 0 with qa < b. Then b C qa > 0 and is in S . Therefore in either case S is nonempty. Hence S is a nonempty subset of N [ ¹0º and therefore has a least element r. If r ¤ 0 we must show that 0 < r < a. Suppose r  a, then r D a C x with x  0 and x < r since a > 0. Then b  qa D r D a C x ) b  .q C 1/a D x. This means that x 2 S. Since x < r this contradicts the minimality of r which is a contradiction. Therefore if r ¤ 0 it follows that 0 < r < a. The only thing left is to show the uniqueness of q and r. Suppose b D q1 a C r1 also. By the construction above r1 must also be the minimal element of S . Hence r1  r and r  r1 so r D r1 . Now b  qa D b  q1 a H) .q1  q/a D 0 but since a > 0 it follows that q1  q D 0 so that q D q1 .

Section 3.1 The Fundamental Theorem of Arithmetic

31

The next idea that is necessary is the concept of greatest common divisor. Definition 3.1.5. Given nonzero integers a; b their greatest common divisor or GCD d > 0 is a positive integer which is a common divisor, that is d ja and d jb, and if d1 is any other common divisor then d1 jd . We denote the greatest common divisor of a; b by either gcd.a; b/ or .a; b/. Certainly, if a; b are nonzero integers with a > 0 and ajb then a D gcd.a; b/. The next result says that given any nonzero integers they do have a greatest common divisor and it is unique. Theorem 3.1.6. Given nonzero integers a; b their GCD exists, is unique and can be characterized as the least positive linear combination of a and b. Proof. Given nonzero a; b consider the set S D ¹ax C by > 0 W x; y 2 Zº: Now a2 C b 2 > 0 so S is a nonempty subset of N and hence has a least element d > 0. We show that d is the GCD. First we must show that d is a common divisor. Now d D ax C by and is the least such positive linear combination. By the division algorithm a D qd C r with 0  r < d . Suppose r ¤ 0. Then r D a  qd D a  q.ax C by/ D .1  qx/a  qby > 0. Hence r is a positive linear combination of a and b and therefore is in S. But then r < d contradicting the minimality of d in S. It follows that r D 0 and so a D qd and d ja. An identical argument shows that d jb and so d is a common divisor of a and b. Let d1 be any other common divisor of a and b. Then d1 divides any linear combination of a and b and so d1 jd . Therefore d is the GCD of a and b. Finally we must show that d is unique. Suppose d1 is another GCD of a and b. Then d1 > 0 and d1 is a common divisor of a; b. Then d1 jd since d is a GCD. Identically d jd1 since d1 is a GCD. Therefore d D ˙d1 and then d D d1 since they are both positive. If .a; b/ D 1 then we say that a; b are relatively prime. It follows that a and b are relatively prime if and only if 1 is expressible as a linear combination of a and b. We need the following three results. Lemma 3.1.7. If d D .a; b/ then a D a1 d and b D b1 d with .a1 ; b1 / D 1. Proof. If d D .a; b/ then d ja and d jb. Hence a D a1 d and b D b1 d . We have d D ax C by D a1 dx C b1 dy:

32

Chapter 3 Prime Elements and Unique Factorization Domains

Dividing both sides of the equation by d we obtain 1 D a1 x C b1 y: Therefore .a1 ; b1 / D 1. Lemma 3.1.8. For any integer c we have that .a; b/ D .a; b C ac/. Proof. Suppose .a; b/ D d and .a; b C ac/ D d1 . Now d is the least positive linear combination of a and b. Suppose d D axCby. d1 is a linear combination of a; bCac so that d1 D ar C .b C ac/s D a.cs C r/ C bs: Hence d1 is also a linear combination of a and b and therefore d1  d . On the other hand d1 ja and d1 j.b C ac/ and so d1 jb. Therefore d1 jd so d1  d . Combining these we must have d1 D d . The next result, called the Euclidean algorithm, provides a technique for both finding the GCD of two integers and expressing the GCD as a linear combination. Theorem 3.1.9 (Euclidean algorithm). Given integers b and a > 0 with a − b, form the repeated divisions b D q1 a C r1 ;

0 < r1 < a

a D q2 r1 C r2 ;

0 < r2 < r1

:: : rn2 D qn rn1 C rn ;

0 < rn < rn1

rn1 D qnC1 rn : The last nonzero remainder rn is the GCD of a; b. Further rn can be expressed as a linear combination of a and b by successively eliminating the ri ’s in the intermediate equations. Proof. In taking the successive divisions as outlined in the statement of the theorem each remainder ri gets strictly smaller and still nonnegative. Hence it must finally end with a zero remainder. Therefore there is a last nonzero remainder rn . We must show that this is the GCD. Now from Lemma 3.1.7 the gcd .a; b/ D .a; b q1 a/ D .a; r1 / D .r1 ; aq2 r1 / D .r1 ; r2 /. Continuing in this manner we have then that .a; b/ D .rn1 ; rn / D rn since rn divides rn1 . This shows that rn is the GCD. To express rn as a linear combination of a and b notice first that rn D rn2  qn rn1 :

Section 3.1 The Fundamental Theorem of Arithmetic

33

Substituting this in the immediately preceding division we get rn D rn2  qn .rn3  qn1 rn2 / D .1 C qn qn1 /rn2  qn rn3 : Doing this successively we ultimately express rn as a linear combination of a and b.

Example 3.1.10. Find the GCD of 270 and 2412 and express it as a linear combination of 270 and 2412. We apply the Euclidean algorithm 2412 D .8/.270/ C 252 270 D .1/.252/ C 18 252 D .14/.18/: Therefore the last nonzero remainder is 18 which is the GCD. We now must express 18 as a linear combination of 270 and 2412. From the first equation 252 D 2412  .8/.270/ which gives in the second equation 270 D 2412  .8/.270/ C 18 H) 18 D .1/.2412/ C .9/.270/ which is the desired linear combination. The next result that we need is Euclid’s lemma. We stated and proved this in the last chapter but we restate it here. Lemma 3.1.11 (Euclid’s lemma). If p is a prime and pjab then pja or pjb. We can now prove the fundamental theorem of arithmetic. Induction suffices to show that there always exists such a decomposition into prime factors. Lemma 3.1.12. Any integer n > 1 can be expressed as a product of primes, perhaps with only one factor. Proof. The proof is by induction. n D 2 is prime so its true at the lowest level. Suppose that any integer 2  k < n can be decomposed into prime factors, we must show that n then also has a prime factorization. If n is prime then we are done. Suppose then that n is composite. Hence n D m1 m2 with 1 < m1 < n, 1 < m2 < n. By the inductive hypothesis both m1 and m2 can be expressed as products of primes. Therefore n can, also using the primes from m1 and m2 , completing the proof.

34

Chapter 3 Prime Elements and Unique Factorization Domains

Before we continue to the fundamental theorem we mention that the existence of a prime decomposition, unique or otherwise, can be used to prove that the set of primes is infinite. The proof we give goes back to Euclid and is quite straightforward. Theorem 3.1.13. There are infinitely many primes. Proof. Suppose that there are only finitely many primes p1 ; : : : ; pn . Each of these is positive so we can form the positive integer N D p1 p2    pn C 1: From Lemma 3.1.12 N has a prime decomposition. In particular there is a prime p which divides N . Then pj.p1 p2    pn C 1/: Since the only primes are assumed p1 ; p2 ; : : : ; pn it follows that p D pi for some i D 1; : : : ; n. But then pjp1 p2    pi : : : pn c so p cannot divide p1    pn C 1 which is a contradiction. Therefore p is not one of the given primes showing that the list of primes must be endless. We can now prove the fundamental theorem of arithmetic. Proof. We assume that n  1. If n  1 we use c D n and the proof is the same. The statement certainly holds for n D 1 with k D 0. Now suppose n > 1. From Lemma 3.1.12, n has a prime decomposition n D p1 p2    pm : We must show that this is unique up to the ordering of the factors. Suppose then that n has another such factorization n D q1 q2    qk with the qi all prime. We must show that m D k and that the primes are the same. Now we have n D p1 p2    pm D q1    qk : Assume that k  m. From n D p1 p2    pm D q1    qk it follows that p1 jq1 q2    qk . From Lemma 3.1.11 then we must have that p1 jqi for some i. But qi is prime and p1 > 1 so it follows that p1 D qi . Therefore we can eliminate p1 and qi from both sides of the factorization to obtain p2    pm D q1    qi 1 qi C1    qk : Continuing in this manner we can eliminate all the pi from the left side of the factorization to obtain 1 D qmC1    qk : If qmC1 ; : : : ; qk were primes this would be impossible. Therefore m D k and each prime pi was included in the primes q1 ; : : : ; qm . Therefore the factorizations differ only in the order of the factors, proving the theorem.

Section 3.2 Prime Elements, Units and Irreducibles

3.2

35

Prime Elements, Units and Irreducibles

We now let R be an arbitrary integral domain and attempt to mimic the divisibility definitions and properties. Definition 3.2.1. Let R be an integral domain. (1) Suppose that a; b 2 R. Then a is a factor or divisor of b if there exists a c 2 R with b D ac. We denote this, as in the integers, by ajb. If a is a factor of b then b is called a multiple of a. (2) An element a 2 R is a unit if a has a multiplicative inverse within R, that is there exists an element a1 2 R with aa1 D 1. (3) A prime element of R is an element p ¤ 0 such that p is not a unit and if pjab then pja or pjb. (4) An irreducible in R is an element c ¤ 0 such that c is not a unit and if c D ab then a or b must be a unit. (5) a and b in R are associates if there exists a unit e 2 R with a D eb. Notice that in the integers Z the units are just ˙1. The set of prime elements coincides with the set of irreducible elements. In Z this are precisely the set of prime numbers. On the other hand if K is a field every nonzero element is a unit so in K there are no prime elements and no irreducible elements. Recall that the modular rings Zn are fields (and integral domains) when n is a prime. In general if n is not a prime then Zn is a commutative ring with an identity and a unit is still an invertible element. We can characterize the units within Zn . Lemma 3.2.2. a 2 Zn is a unit if and only if .a; n/ D 1. Proof. Suppose .a; n/ D 1. Then there exist x; y 2 Z such that ax C ny D 1. This implies that ax  1 mod n which in turn implies that ax D 1 in Zn and therefore a is a unit. Conversely suppose a is a unit in Zn . Then there is an x 2 Zn with ax D 1. In terms of congruence then ax  1

mod n H) njax  1 H) ax  1 D ny H) ax  ny D 1:

Therefore 1 is a linear combination of a and n and so .a; n/ D 1. If R is an integral domain then the set of units within R will form a group. Lemma 3.2.3. If R is a commutative ring with an identity then the set of units in R form an abelian group under ring multiplication. This is called the unit group of R denoted U.R/.

36

Chapter 3 Prime Elements and Unique Factorization Domains

Proof. The commutativity and associativity of U.R/ follow from the ring properties. The identity of U.R/ is the multiplicative identity of R while the ring multiplicative inverse for each unit is the group inverse. We must show that U.R/ is closed under ring multiplication. If a 2 R is a unit we denote its multiplicative inverse by a1 . Now suppose a; b 2 U.R/. Then a1 ; b 1 exist. It follows that .ab/.b 1 a1 / D a.bb 1 /a1 D aa1 D 1: Hence ab has an inverse, namely b 1 a1 (D a1 b 1 in a commutative ring) and hence ab is also a unit. Therefore U.R/ is closed under ring multiplication. In general irreducible elements are not prime. Consider for example the subring of the complex numbers (see exercises) given by p p R D ZŒi 5 D ¹x C i 5y W x; y 2 Zº: This is a subring of the complex numbers C and hence can have no zero divisors. Therefore R is an integralpdomain. For an element x C iy 5 2 R define its norm by p p N.x C iy 5/ D jx C iy 5j D x 2 C 5y 2 : Since x; y 2 Z it is clear that the norm of an element in R is a nonnegative integer. Further if a 2 R with N.a/ D 0 then a D 0. We have the following result concerning the norm. Lemma 3.2.4. Let R and N be as above. Then (1) N.ab/ D N.a/N.b/ for any elements a; b 2 R. (2) The units of R are those a 2 R with N.a/ D 1. In R the only units are ˙1. Proof. The fact that the norm is multiplicative is straightforward and left to the exercises. If a 2 R is a unit then there exists a multiplicative inverse b 2 R with ab D 1. Then N.ab/ D N.a/N.b/ D 1. Since both N.a/ and N.b/ are nonnegative integers we must have N.a/ D N.b/ D 1. p Conversely suppose that N.a/ D 1. If a D x C iy 5 then x 2 C 5y 2 D 1. Since x; y 2 Z we must have y D 0 and x 2 D 1. Then a D x D ˙1. Using this lemma we can show that R possesses irreducible elements that are not prime. p Lemma 3.2.5. Let R be as above. Then 3 D 3 C i 0 5 is an irreducible element in R but 3 is not prime.

Section 3.2 Prime Elements, Units and Irreducibles

37

Proof. Suppose that 3 D ab with a; b 2 R and a; b nonunits. Then N.3/ D 9 D N.a/N.b/ with p neither N.a/ D 1 nor N.b/ D 1. Hence both N.a/ D 3 and N.b/ D 3. Let a D x C iy 5. It follows that x 2 C 5y 2 D 3. Since x; y 2 Z this is impossible. Therefore one of a or b must be a unit and 3 is an irreducible element. p p We show that 3 is not prime in R. Let a D 2 C i 5 and b D 2  i 5. Then ab D 9 and hence 3jab. Suppose 3ja so that a D 3c for some c 2 R. Then 9 D N.a/ D N.3/N.c/ D 9N.c/ H) N.c/ D 1: Therefore c is a unit in R and from Lemma 3.2.4 we get c D ˙1. Hence a D ˙3. This is a contradiction, so 3 does not divide a. An identical argument shows that 3 does not divide b. Therefore 3 is not a prime element in R. We now examine the relationship between prime elements and irreducibles. Theorem 3.2.6. Let R be an integral domain. Then (1) Each prime element of R is irreducible. (2) p 2 R is a prime element if and only if p ¤ 0 and hpi D pR is a prime ideal. (3) p 2 R is irreducible if and only if p ¤ 0 and hpi D pR is maximal in the set of all principal ideals of R which are not equal to R. Proof. (1) Suppose that p 2 R is a prime element and p D ab. We must show that either a or b must be a unit. Now pjab so either pja or pjb. Without loss of generality we may assume that pja, so a D pr for some r 2 R. Hence p D ab D .pr/b D p.rb/. However R is an integral domain so p  prb D p.1  rb/ D 0 implies that 1  rb D 0 and hence rb D 1. Therefore b is a unit and hence p is irreducible. (2) Suppose that p is a prime element. Then p ¤ 0. Consider the ideal pR and suppose that ab 2 pR. Then ab is a multiple of p and hence pjab. Since p is prime it follows that pja or pjb. If pja then a 2 pR while if pjb then b 2 pR. Therefore pR is a prime ideal. Conversely suppose that pR is a prime ideal and suppose that p D ab. Then ab 2 pR so a 2 pR or b 2 pR. If a 2 pR then pja and if b 2 pR then pjb and so p is prime. (3) Let p be irreducible then p ¤ 0. Suppose that pR  aR where a 2 R. Then p D ra for some r 2 R. Since p is irreducible it follows that either a is a unit or r is a unit. If r is a unit we have pR D raR D aR ¤ R since p is not a unit. If a is a unit then aR D R and pR D rR ¤ R. Therefore pR is maximal in the set of principal ideals not equal to R. Conversely suppose p ¤ 0 and pR is a maximal ideal in the set of principal ideals ¤ R. Let p D ab with a not a unit. We must show that b is a unit. Since aR ¤ R and pR  aR from the maximality we must have pR D aR. Hence a D rp for some r 2 R. Then p D ab D rpb and as before we must have rb D 1 and b a unit.

38

Chapter 3 Prime Elements and Unique Factorization Domains

Theorem 3.2.7. Let R be a principle ideal domain. Then: (1) An element p 2 R is irreducible if and only if it is a prime element. (2) A nonzero ideal of R is a maximal ideal if and only if it is a prime ideal. (3) The maximal ideals of R are precisely those ideals pR where p is a prime element. Proof. First note that ¹0º is a prime ideal but not maximal. (1) We already know that prime elements are irreducible. To show the converse suppose that p is irreducible. Since R is a principal ideal domain from Theorem 3.2.6 we have that pR is a maximal ideal, and each maximal ideal is also a prime ideal. Therefore from Theorem 3.2.6 we have that p is a prime element. (2) We already know that each maximal ideal is a prime ideal. To show the converse suppose that I ¤ ¹0º is a prime ideal. Then I D pR where p is a prime element with p ¤ 0. Therefore p is irreducible from part (1) and hence pR is a maximal ideal from Theorem 3.2.6. (3) This follows directly from the proof in part (2) and Theorem 3.2.6.

3.3

Unique Factorization Domains

We now consider integral domains where there is unique factorization into primes. If R is an integral domain and a; b 2 R then we say that a and b are associates if there exists a unit  2 R with a D b. Definition 3.3.1. An integral domain D is a unique factorization domain or UFD if for each d 2 D then either d D 0, d is a unit or d has a factorization into primes which is unique up to ordering and unit factors. This means that if r D p1    pm D q1    qk then m D k and each pi is an associate of some qj . There are several relationships in integral domains that are equivalent to unique factorization. Definition 3.3.2. Let R be an integral domain. (1) R has property (A) if and only if for each nonunit a ¤ 0 there are irreducible elements q1 ; : : : ; qr 2 R satisfying a D q1    qr . (2) R has property (A0 ) if and only if for each nonunit a ¤ 0 there are prime elements p1 ; : : : ; pr 2 R satisfying a D p1    pr .

Section 3.3 Unique Factorization Domains

39

(3) R has property (B) if and only if whenever q1 ; : : : ; qr and q10 ; : : : ; qs0 are irreducible elements of R with q1    qr D q10    qs0 then r D s and there is a permutation  2 Sr such that for each i 2 ¹1; : : : ; rº 0 are associates (uniqueness up to ordering and unit the elements qi and q.i/ factors). (4) R has property (C) if and only if each irreducible element of R is a prime element. Notice that properties (A) and (C) together are equivalent to what we defined as unique factorization. Hence an integral domain satisfying (A) and (C) is a UFD. We show next, that there are other equivalent formulations. Theorem 3.3.3. In an integral domain R the following are equivalent: (1) R is a UFD. (2) R satisfies properties (A) and (B). (3) R satisfies properties (A) and (C). (4) R satisfies property (A0 ). Proof. As remarked before the statement of the theorem by definition (A) and (C) are equivalent to unique factorization. We show here that (2), (3) and (4) are equivalent. First we show that (2) implies (3). Suppose that R satisfies properties (A) and (B). We must show that it also satisfies (C), that is we must show that if q 2 R is irreducible then q is prime. Suppose that q 2 R is irreducible and qjab with a; b 2 R. Then we have ab D cq for some c 2 R. If a is a unit from ab D cq we get that b D a1 cq and qjb. Identically if b is a unit. Therefore we may assume that neither a nor b are units. If c D 0 then since R is an integral domain either a D 0 or b D 0 and qja or qjb. We may assume then that c ¤ 0. If c is a unit then q D c 1 ab and since q is irreducible either c 1 a or b are units. If c 1 a is a unit then a is also a unit so if c is a unit either a or b are units contrary to our assumption. Therefore we may assume that c ¤ 0 and c is not a unit. From property (A) we have a D q1    q r b D q10    qs0 c D q100    q t00

40

Chapter 3 Prime Elements and Unique Factorization Domains

where q1 ; : : : qr ; q10 ; : : : ; qs0 ; q100 ; : : : q t00 are all irreducibles. Hence q1    qr q10    qs0 D q100    q t00  q: From property (B) q is an associate of some qi or qj0 . Hence qjqi or qjqj0 . It follows that qja or qjb and therefore q is a prime element. That (3) implies (4) is direct. We show that (4) implies (2). Suppose that R satisfies property (A0 ). We must show that it satisfies both (A) and (B). We show first that (A) follows from (A0 ) by showing that irreducible elements are prime. Suppose that q is irreducible. Then from (A0 ) we have q D p1    pr with each pi prime. It follows without loss of generality that p2    pr is a unit and p1 is a nonunit and hence pi j1 for i D 2; : : : ; r. Thus q D p1 and q is prime. Therefore (A) holds. We now show that (B) holds. Let q1    qr D q10    qs0 where qi ; qj0 are all irreducibles and hence primes. Then q10 jq1    qr and so q10 jqi for some i . Without loss of generality suppose q10 jq1 . Then q1 D aq10 . Since q1 is irreducible it follows that a is a unit and q1 and q10 are associates. It follows then that aq2    qr D q20  qs0 since R has no zero divisors. Property (B) holds then by induction and the theorem is proved. Note that in our new terminology Z is a UFD. In the next section we will present other examples of UFD’s however not every integral domain is a unique factorization domain. As we defined in the last section let R be the following subring of C. p p R D ZŒi 5 D ¹x C i 5y W x; y 2 Zº: R is an integral domain and we showed using the norm 3 is an irreducible in R. p that p are also irreducibles Analogously we can show that the elements 2 C i p5; 2  i 5p in R and further 3 is not an associate of either 2 C i 5 or 2  i 5. Then p p 9 D 3  3 D .2 C i 5/.2  i 5/

41

Section 3.4 Principal Ideal Domains and Unique Factorization

give two different decompositions for an element in terms of irreducible elements. The fact that R is not a UFD also follows from the fact that 3 is an irreducible element which is not prime. Notice also that Euclid’s proof, that there are infinitely many primes in Z, works in any UFD. Theorem 3.3.4. Let R be a UFD. Then R has infinitely many prime elements. Unique factorization is tied to the famous solution of Fermat’s big theorem. Wiles and Taylor in 1995 proved the following. Theorem 3.3.5. The equation x p C y p D z p has no integral solutions with xyz ¤ 0 for any prime p  3. Kummer tried to prove this theorem by attempting to factor x p D z p y p . Call the 2 i conclusion of Theorem 3.3.4 in an integral domain R property .Fp /. Let  D e p . Then p1 Y p p .z   j y/: z y D j D0

View this equation in the ring R D ZΠD

² p1 X

³ j

aj  W aj 2 Z :

j D0

Kummer proved that if R is a UFD then property .Fp / holds. However, independently, from Uchida and Montgomery (1971) R is a UFD only if p  19 (see [41]).

3.4

Principal Ideal Domains and Unique Factorization

In this section we prove that every principal ideal domain (PID) is a unique factorization domain (UFD). We say that an ascending chain of ideals in R I1  I2      In     becomes stationary if there exists an m such that Ir D Im for all r  m. Theorem 3.4.1. Let R be an integral domain. If each ascending chain of principal ideals in R becomes stationary, then R satisfies property (A). Proof. Suppose that a ¤ 0 is a not a unit in R. Suppose that a is not a product of irreducible elements. Clearly then a cannot itself be irreducible. Hence a D a1 b1 with a1 ; b1 2 R and a1 ; b1 are not units. If both a1 or b1 can be expressed as a product

42

Chapter 3 Prime Elements and Unique Factorization Domains

of irreducible elements then so can a. Without loss of generality then suppose that a1 is not a product of irreducible elements. Since a1 ja we have the inclusion of ideals aR a1 R. If a1 R D aR then a1 2 aR and a1 D ar D a1 b1 r which implies that b1 is a unit contrary to our assumption. Therefore aR ¤ a1 R and the inclusion is proper. By iteration then we obtain a strictly increasing chain of ideals aR  a1 R      an R     : From our hypothesis on R this must become stationary contradicting the argument above that the inclusion is proper. Therefore a must be a product of irreducibles. Theorem 3.4.2. Each principal ideal domain R is a unique factorization domain. Proof. Suppose that R is a principal ideal domain. R satisfies property (C) by Theorem 3.2.7(1), so to show that it is a unique factorization domain we must show that it also satisfies property (A). From the previous theorem it suffices to show that each ascending chain of principal ideals becomes stationary. Consider such an ascending chain a1 R  a2 R      an R     : Now let I D

1 [

ai R:

i D1

Now I is an ideal in R and hence a principal ideal. Therefore I D aR for some a 2 R. Since I is a union there exists an m such that a 2 am R. Therefore I D aR  am R and hence I D am R and ai R  am R for all i  m. Therefore the chain becomes stationary and from Theorem 3.4.1 R satisfies property (A). Since we showed that the integers Z are a PID we can recover the fundamental theorem of arithmetic from Theorem 3.4.2. We now present another important example of a PID and hence a UFD. In the next chapter we will look in detail at polynomials with coefficients in an integral domain. Below we consider polynomials with coefficients in a field and for the present leave out many of the details. If F is a field and n is a nonnegative integer, then a polynomial of degree n over F is a formal sum of the form P .x/ D a0 C a1 x C    C an x n with ai 2 F for i D 0; : : : ; n, an ¤ 0, and x an indeterminate. A polynomial P .x/ over F is either a polynomial of some degree or the expression P .x/ D 0, which is called the zero polynomial and has degree 1. We denote the degree of P .x/ by deg P .x/. A polynomial of zero degree has the form P .x/ D a0 and is

Section 3.4 Principal Ideal Domains and Unique Factorization

43

called a constant polynomial and can be identified with the corresponding element of F . The elements ai 2 F are called the coefficients of P .x/; an is the leading coefficient. If an D 1, P .x/ is called a monic polynomial. Two nonzero polynomials are equal if and only if they have the same degree and exactly the same coefficients. A polynomial of degree 1 is called a linear polynomial while one of degree two is a quadratic polynomial. We denote by F Œx the set of all polynomials over F and we will show that F Œx becomes a principal ideal domain and hence a unique factorization domain. We first define addition, subtraction, and multiplication on F Œx by algebraic manipulation. That is, suppose P .x/ D a0 C a1 x C    C an x n ; Q.x/ D b0 C b1 x C    C bm x m then P .x/ ˙ Q.x/ D .a0 ˙ b0 / C .a1 ˙ b1 /x C    ; that is, the coefficient of x i in P .x/ ˙ Q.x/ is ai ˙ bi , where ai D 0 for i > n and bj D 0 for j > m. Multiplication is given by P .x/Q.x/ D .a0 b0 /C.a1 b0 Ca0 b1 /xC.a0 b2 Ca1 b1 Ca2 b0 /x 2 C  C.an bm /x nCm ; that is, the coefficient of x i in P .x/Q.x/ is .a0 bi C a1 bi 1 C    C ai b0 /. Example 3.4.3. Let P .x/ D 3x 2 C 4x  6 and Q.x/ D 2x C 7 be in QŒx. Then P .x/ C Q.x/ D 3x 2 C 6x C 1 and P .x/Q.x/ D .3x 2 C 4x  6/.2x C 7/ D 6x 3 C 29x 2 C 16x  42: From the definitions the following degree relationships are clear. The proofs are in the exercises. Lemma 3.4.4. Let 0 ¤ P .x/; 0 ¤ Q.x/ in F Œx. Then: (1) deg P .x/Q.x/ D deg P .x/ C deg Q.x/. (2) deg.P .x/ ˙ Q.x//  Max.deg P .x/; deg Q.x// if P .x/ ˙ Q.x/ ¤ 0. We next obtain the following. Theorem 3.4.5. If F is a field, then F Œx forms an integral domain. F can be naturally embedded into F Œx by identifying each element of F with the corresponding constant polynomial. The only units in F Œx are the nonzero elements of F . Proof. Verification of the basic ring properties is solely computational and is left to the exercises. Since deg P .x/Q.x/ D deg P .x/ C deg Q.x/, it follows that if neither P .x/ ¤ 0 nor Q.x/ ¤ 0 then P .x/Q.x/ ¤ 0 and therefore F Œx is an integral domain.

44

Chapter 3 Prime Elements and Unique Factorization Domains

If G.x/ is a unit in F Œx, then there exists an H.x/ 2 F Œx with G.x/H.x/ D 1. From the degrees we have deg G.x/ C deg H.x/ D 0 and since deg G.x/  0, deg H.x/  0. This is possible only if deg G.x/ D deg H.x/ D 0. Therefore G.x/ 2 F . Now that we have F Œx as an integral domain we proceed to show that F Œx is a principal ideal domain and hence there is unique factorization into primes. We first repeat the definition of a prime in F Œx. If 0 ¤ f .x/ has no nontrivial, nonunit factors (it cannot be factorized into polynomials of lower degree) then f .x/ is a prime in F Œx or a prime polynomial. A prime polynomial is also called an irreducible polynomial. Clearly, if deg g.x/ D 1 then g.x/ is irreducible. The fact that F Œx is a principal ideal domain follows from the division algorithm for polynomials, which is entirely analogous to the division algorithm for integers. Lemma 3.4.6 (division algorithm in F Œx). If 0 ¤ f .x/, 0 ¤ g.x/ 2 F Œx then there exist unique polynomials q.x/; r.x/ 2 F Œx such that f .x/ D q.x/g.x/ C r.x/ where r.x/ D 0 or deg r.x/ < deg g.x/. (The polynomials q.x/ and r.x/ are called respectively the quotient and remainder.) This theorem is essentially long division of polynomials. A formal proof is based on induction on the degree of g.x/. We omit this but give some examples from QŒx. Example 3.4.7.

(a) Let f .x/ D 3x 4  6x 2 C 8x  6, g.x/ D 2x 2 C 4. Then

3x 4  6x 2 C 8x  6 3 D x 2  6 with remainder 8x C 18: 2 2x C 4 2 Thus here q.x/ D 32 x 2  6, r.x/ D 8x C 18. (b) Let f .x/ D 2x 5 C 2x 4 C 6x 3 C 10x 2 C 4x, g.x/ D x 2 C x. Then 2x 5 C 2x 4 C 6x 3 C 10x 2 C 4x D 2x 3 C 6x C 4: x2 C x Thus here q.x/ D 2x 3 C 6x C 4 and r.x/ D 0. Theorem 3.4.8. Let F be a field. Then the polynomial ring F Œx is a principal ideal domain and hence a unique factorization domain. Proof. The proof is essentially analogous to the proof in the integers. Let I be an ideal in F Œx with I ¤ F Œx. Let f .x/ be a polynomial in I of minimal degree. We claim that I D hf .x/i the principal ideal generated by f .x/. Let g.x/ 2 I . We must show that g.x/ is a multiple of f .x/. By the division algorithm in F Œx we have g.x/ D q.x/f .x/ C r.x/

45

Section 3.5 Euclidean Domains

where r.x/ D 0 or deg.r.x// < deg.f .x//. If r.x/ ¤ 0 then deg.r.x// < deg.f .x//. However r.x/ D g.x/  q.x/f .x/ 2 I since I is an ideal and g.x/; f .x/ 2 I . This is a contradiction since f .x/ was assumed to be a polynomial in I of minimal degree. Therefore r.x/ D 0 and hence g.x/ D q.x/f .x/ is a multiple of f .x/. Therefore each element of I is a multiple of f .x/ and hence I D hf .x/i. Therefore F Œx is a principal ideal domain and from Theorem 3.4.2 a unique factorization domain. We proved that in a principal ideal domain every ascending chain of ideals becomes stationary. In general a ring R (commutative or not) satisfies the ascending chain condition or ACC if every ascending chain of left (or right) ideals in R becomes stationary. A ring satisfying the ACC is called a Noetherian ring.

3.5

Euclidean Domains

In analyzing the proof of unique factorization in both Z and F Œx, it is clear that it depends primarily on the division algorithm. In Z the division algorithm depended on the fact that the positive integers could be ordered and in F Œx on the fact that the degrees of nonzero polynomials are nonnegative integers and hence could be ordered. This basic idea can be generalized in the following way. Definition 3.5.1. An integral domain D is a Euclidean domain if there exists a function N from D ? D D n ¹0º to the nonnegative integers such that (1) N.r1 /  N.r1 r2 / for any r1 ; r2 2 D ? . (2) For all r1 ; r2 2 D with r1 ¤ 0 there exist q; r 2 D such that r2 D qr1 C r where either r D 0 or N.r/ < N.r1 /. The function N is called a Euclidean norm on D. Therefore Euclidean domains are precisely those integral domains which allow division algorithms. In the integers Z define N.z/ D jzj. Then N is a Euclidean norm on Z and hence Z is a Euclidean domain. On F Œx define N.p.x// D deg.p.x// if p.x/ ¤ 0. Then N is also a Euclidean norm on F Œx so that F Œx is also a Euclidean domain. In any Euclidean domain we can mimic the proofs of unique factorization in both Z and F Œx to obtain the following: Theorem 3.5.2. Every Euclidean domain is a principal ideal domain and hence a unique factorization domain. Before proving this theorem we must develop some results on the number theory of general Euclidean domains. First some properties of the norm.

46

Chapter 3 Prime Elements and Unique Factorization Domains

Lemma 3.5.3. If R is a Euclidean domain then (a) N.1/ is minimal among ¹N.r/ W r 2 R? º. (b) N.u/ D N.1/ if and only if u is a unit. (c) N.a/ D N.b/ for a; b 2 R? if a; b are associates. (d) N.a/ < N.ab/ unless b is a unit. Proof. (a) From property (1) of Euclidean norms we have N.1/  N.1  r/ D N.r/

for any r 2 R? :

(b) Suppose u is a unit. Then there exists u1 with u  u1 D 1. Then N.u/  N.u  u1 / D N.1/: From the minimality of N.1/ it follows that N.u/ D N.1/. Conversely suppose N.u/ D N.1/. Apply the division algorithm to get 1 D qu C r: If r ¤ 0 then N.r/ < N.u/ D N.1/ contradicting the minimality of N.1/. Therefore r D 0 and 1 D qu. Then u has a multiplicative inverse and hence is a unit. (c) Suppose a; b 2 R? are associates. Then a D ub with u a unit. Then N.b/  N.ub/ D N.a/: On the other hand b D u1 a so N.a/  N.u1 a/ D N.b/: Since N.a/  N.b/ and N.b/  N.a/ it follows that N.a/ D N.b/. (d) Suppose N.a/ D N.ab/. Apply the division algorithm a D q.ab/ C r where r D 0 or N.r/ < N.ab/. If r ¤ 0 then r D a  qab D a.1  qb/ H) N.ab/ D N.a/  N.a.1  qb// D N.r/ contradicting that N.r/ < N.ab/. Hence r D 0 and a D q.ab/ D .qb/a. Then a D .qb/a D 1  a H) qb D 1 since there are no zero divisors in an integral domain. Hence b is a unit. Since N.a/  N.ab/ it follows that if b is not a unit we must have N.a/ < N.ab/.

47

Section 3.5 Euclidean Domains

We can now prove Theorem 3.5.2. Proof. Let D be a Euclidean domain. We show that each ideal I ¤ D in D is principal. Let I ¤ D be an ideal in D. If I D ¹0º then I D h0i and I is principal. Therefore we may assume that there are nonzero elements in I . Hence there are elements x 2 I with strictly positive norm. Let a be an element of I of minimal norm. We claim that I D hai. Let b 2 I . We must show that b is a multiple of a. Now by the division algorithm b D qa C r where either r D 0 or N.r/ < N.a/. As in Z and F Œx we have a contradiction if r ¤ 0. In this case N.r/ < N.a/ but r D b  qa 2 I since I is an ideal contradicting the minimality of N.a/. Therefore r D 0 and b D qa and hence I D hai. As a final example of a Euclidean domain we consider the Gaussian integers ZŒi D ¹a C bi W a; b 2 Zº: It was first observed by Gauss that this set permits unique factorization. To show this we need a Euclidean norm on ZŒi. Definition 3.5.4. If z D a C bi 2 ZŒi then its norm N.z/ is defined by N.a C bi / D a2 C b 2 : The basic properties of this norm follow directly from the definition (see exercises). Lemma 3.5.5. If ˛; ˇ 2 ZŒi then: (1) N.˛/ is an integer for all ˛ 2 ZŒi. (2) N.˛/  0 for all ˛ 2 ZŒi. (3) N.˛/ D 0 if and only if ˛ D 0. (4) N.˛/  1 for all ˛ ¤ 0. (5) N.˛ˇ/ D N.˛/N.ˇ/, that is the norm is multiplicative. From the multiplicativity of the norm we have the following concerning primes and units in ZŒi. Lemma 3.5.6.

(1) u 2 ZŒi is a unit if and only if N.u/ D 1.

(2) If  2 ZŒi and N./ D p where p is an ordinary prime in Z then  is a prime in ZŒi.

48

Chapter 3 Prime Elements and Unique Factorization Domains

Proof. Certainly u is a unit if and only if N.u/ D N.1/. But in ZŒi we have N.1/ D 1 so the first part follows. Suppose next that  2 ZŒi with N./ D p for some p 2 Z. Suppose that  D 1 2 . From the multiplicativity of the norm we have N./ D p D N.1 /N.2 /: Since each norm is a positive ordinary integer and p is a prime it follows that either N.1 / D 1 or N.2 / D 1. Hence either 1 or 2 is a unit. Therefore  is a prime in ZŒi. Armed with this norm we can show that ZŒi is a Euclidean domain. Theorem 3.5.7. The Gaussian integers ZŒi form a Euclidean domain. Proof. That ZŒi forms a commutative ring with an identity can be verified directly and easily. If ˛ˇ D 0 then N.˛/N.ˇ/ D 0 and since there are no zero divisors in Z we must have N.˛/ D 0 or N.ˇ/ D 0. But then either ˛ D 0 or ˇ D 0 and hence ZŒi is an integral domain. To complete the proof we show that the norm N is a Euclidean norm. From the multiplicativity of the norm we have if ˛; ˇ ¤ 0 N.˛ˇ/ D N.˛/N.ˇ/  N.˛/

since N.ˇ/  1:

Therefore property (1) of Euclidean norms is satisfied. We must now show that the division algorithm holds. Let ˛ D a C bi and ˇ D c C d i be Gaussian integers. Recall that for a nonzero complex number z D x C iy its inverse is z x  iy 1 : D 2 D 2 z jzj x C y2 Therefore as a complex number ˇ c  di ˛ D ˛ 2 D .a C bi / 2 ˇ jˇj c C d2 ac C bd ac  bd D 2 C 2 i D u C iv: 2 c Cd c C d2 Now since a; b; c; d are integers u; v must be rationals. The set ¹u C iv W u; v 2 Qº is called the set of the Gaussian rationals.

Section 3.5 Euclidean Domains

49

If u; v 2 Z then u C iv 2 ZŒi, ˛ D qˇ with q D u C iv, and we are done. Otherwise choose ordinary integers m; n satisfying ju  mj  12 and jv  nj  12 and let q D m C i n. Then q 2 ZŒi. Let r D ˛  qˇ. We must show that N.r/ < N.ˇ/. Working with complex absolute value we get ˇ ˇ ˇ ˇ˛ jrj D j˛  qˇj D jˇjˇˇ  q ˇˇ: ˇ Now

s ˇ ˇ  2  2 q ˇ ˇ˛ 1 1 ˇ  q ˇ D j.u  m/ C i.v  n/j D .u  m/2 C .v  n/2  C < 1: ˇ ˇˇ 2 2 Therefore jrj < jˇj H) jrj2 < jˇj2 H) N.r/ < N.ˇ/ completing the proof. Since ZŒi forms a Euclidean domain it follows from our previous results that ZŒi must be a principal ideal domain and hence a unique factorization domain. Corollary 3.5.8. The Gaussian integers are a UFD. Since we will now be dealing with many kinds of integers we will refer to the ordinary integers Z as the rational integers and the ordinary primes p as the rational primes. It is clear that Z can be embedded into ZŒi. However not every rational prime is also prime in ZŒi. The primes in ZŒi are called the Gaussian primes. For example we can show that both 1 C i and 1  i are Gaussian primes, that is primes in ZŒi. However .1 C i/.1  i / D 2 so that the rational prime 2 is not a prime in ZŒi. Using the multiplicativity of the Euclidean norm in ZŒi we can describe all the units and primes in ZŒi. Theorem 3.5.9.

(1) The only units in ZŒi are ˙1; ˙i.

(2) Suppose  is a Gaussian prime. Then  is either: (a) a positive rational prime p  3 mod 4 or an associate of such a rational prime. (b) 1 C i or an associate of 1 C i. (c) a C bi or a  bi where a > 0, b > 0, a is even and N./ D a 2 C b 2 D p with p a rational prime congruent to 1 mod 4 or an associate of a C bi or a  bi . Proof. (1) Suppose u D x C iy 2 ZŒi is a unit. Then from Lemma 3.5.6 we have N.u/ D x 2 C y 2 D 1 implying that .x; y/ D .0; ˙1/ or .x; y/ D .˙1; 0/. Hence u D ˙1 or u D ˙i.

50

Chapter 3 Prime Elements and Unique Factorization Domains

(2) Now suppose that  is a Gaussian prime. Since N./ D  and  2 ZŒi it follows that jN./. N./ is a rational integer so N./ D p1    pk where the pi ’s are rational primes. By Euclid’s lemma jpi for some pi and hence a Gaussian prime must divide at least one rational prime. On the other hand suppose jp and jq where p; q are different primes. Then .p; q/ D 1 and hence there exist x; y 2 Z such that 1 D px C qy. It follows that j1 a contradiction. Therefore a Gaussian prime divides one and only one rational prime. Let p be the rational prime that  divides. Then N./jN.p/ D p 2 . Since N./ is a rational integer it follows that N./ D p or N./ D p2 . If  D a C bi then a2 C b 2 D p or a2 C b 2 D p 2 . If p D 2 then a2 C b 2 D 2 or a2 C b 2 D 4. It follows that  D ˙2; ˙2i or  D 1 C i or an associate of 1 C i . Since .1 C i/.1  i / D 2 and neither 1 C i nor 1  i are units it follows that neither 2 nor any of its associates are primes. Then  D 1 C i or an associate of 1 C i . To see that 1 C i is prime suppose 1 C i D ˛ˇ. Then N.1 C i / D 2 D N.˛/N.ˇ/. It follows that either N.˛/ D 1 or N.ˇ/ D 1 and either ˛ or ˇ is a unit. If p ¤ 2 then either p  3 mod 4 or p  1 mod 4. Suppose first that p  3 mod 4. Then a2 C b 2 D p would imply from Fermat’s two-square theorem (see [35]) that p  1 mod 4. Therefore from the remarks above a2 C b 2 D p 2 and N./ D N.p/. Since jp we have  D ˛p with ˛ 2 ZŒi. From N./ D N.p/ we get that N.˛/ D 1 and ˛ is a unit. Therefore  and p are associates. Hence in this case  is an associate of a rational prime congruent to 3 mod 4. Finally suppose p  1 mod 4. From the remarks above either N./ D p or N./ D p 2 . If N./ D p 2 then a2 C b 2 D p 2 . Since p  1 mod 4 from Fermat’s two square theorem there exist m; n 2 Z with m2 C n2 D p. Let u D m C i n then the norm N.u/ D p. Since p is a rational prime it follows that u is a Gaussian prime. Similarly its conjugate u is also a Gaussian prime. Now uu D p 2 D N./. Since jN./ it follows that juu and from Euclid’s lemma either ju or ju. If ju they are associates since both are primes. But this is a contradiction since N./ ¤ N.u/. The same is true if ju. It follows that if p  1 mod 4, then N./ ¤ p 2 . Therefore in this case N./ D p D a2 C b 2 . An associate of  has both a; b > 0 (see exercises). Further since a2 C b 2 D p one of a or b must be even. If a is odd then b is even then i  is an associate of  with a even completing the proof. Finally we mention that the methods used in ZŒi cannot be applied to all quadratic p integers. For example we have seen that there is not unique factorization in ZŒ 5.

Section 3.6 Overview of Integral Domains

3.6

51

Overview of Integral Domains

Here we present some additional definitions for special types of integral domains. Definition 3.6.1. (1) A Dedekind domain D is an integral domain such that each nonzero proper ideal A (¹0º ¤ A ¤ R) can be written uniquely as a product of prime ideals A D P1    Pr with each Pi a prime ideal and the factorization is unique up to ordering. (2) A Prüfer ring is an integral domain such that A  .B \ C / D AB \ AC for all ideals A; B; C in R. Dedekind domains arise naturally in algebraic number theory. It can be proved that the rings of algebraic integers in any algebraic number field are Dedekind domains (see [35]). If R is a Dedekind domain it is also a Prüfer Ring. If R is a Prüfer ring and a unique factorization domain then R is a principal ideal domain. In the next chapter we will prove a theorem due to Gauss that if R is a UFD then the polynomial ring RŒx is also a UFD. If K is a field we have already seen that KŒx is a UFD. Hence the polynomial ring in several variables KŒx1 ; : : : ; xn  is also a UFD. This fact plays an important role in algebraic geometry.

3.7

Exercises

1. Let R be an integral domain and let  2 R n .U.R/ [ ¹0º/. Show: (i) If for each a 2 R with  − a there exist ;  2 R with  C a D 1 then  is a prime element of R. (ii) Give an example for a prime element  in an UFD R, which does not satisfy the conditions of (i). 2. Let R be an UFD and let a1 ; : : : ; a t be pairwise coprime elements of R. If a1    a t is an m-th power (m 2 N), then all factors ai are an associate of an m-th power. Is each ai necessarily an m-th power? p p p 3, ZŒ 5 and ZŒ 7 3. Decide if the unit pis finite or infinite. For which pgroup of ZŒ p a 2 Z are .1  5/ and .a C 5/ associates in ZŒ 5? p p 4. Let k 2 Z andp k ¤ x 2 for all x 2 Z. Let ˛ D a C b k and ˇ D c C d k be elements of ZŒ k and N.˛/ D a2  kb 2 , N.ˇ/ D c 2  kd 2 . Show:

52

Chapter 3 Prime Elements and Unique Factorization Domains

(i) The equality of the absolutepvalues of N.˛/ and N.ˇ/ is necessary for the association of ˛ and ˇ in ZŒ k. Is this constraint also sufficient? p (ii) Sufficient for the irreducibility of ˛ in ZŒ k is the irreducibility of N.˛/ in Z. Is this also necessary? 5. In general irreducible elements are not prime. Consider the set of complex number given by p p R D ZŒi 5 D ¹x C i 5y W x; y 2 Zº: Show that they form a subring of C. p 6. For an element x C iy 5 2 R define its norm by p p N.x C iy 5/ D jx C iy 5j D x 2 C 5y 2 : Prove that the norm is multiplicative, that is N.ab/ D N.a/N.b/. 7. Prove Lemma 3.4.4. 8. Prove that the set of polynomials RŒx with coefficients in a ring R forms a ring. 9. Prove the basic properties of the norm of the Gaussian integers. If ˛; ˇ 2 ZŒi then: (i) N.˛/ is an integer for all ˛ 2 ZŒi. (ii) N.˛/  0 for all ˛ 2 ZŒi. (iii) N.˛/ D 0 if and only if ˛ D 0. (iv) N.˛/  1 for all ˛ ¤ 0. (v) N.˛ˇ/ D N.˛/N.ˇ/, that is the norm is multiplicative.

Chapter 4

Polynomials and Polynomial Rings

4.1

Polynomials and Polynomial Rings

In the last chapter we saw that if K is a field then the set of polynomials with coefficients in K, which we denoted KŒx, forms a unique factorization domain. In this chapter we take a more detailed look at polynomials over a general ring R. We then prove that if R is a UFD then the polynomial ring RŒx is also a UFD. We first take a formal look at polynomials. Let R be a commutative ring with an identity. Consider the set RQ of functions f from the nonnegative integers N D N [¹0º into R with only a finite number of values nonzero. That is RQ D ¹f W N ! R W f .n/ ¤ 0 for only finitely many nº: On RQ we define the following addition and multiplication .f C g/.m/ D f .m/ C g.m/ X f .i/g.j /: .f  g/.m/ D i Cj Dm

If we let x D .0; 1; 0; : : :/ and identify .r; 0; : : :/ with r 2 R then x 0 D .1; 0; : : :/ D 1 and

x i C1 D x  x i :

Now if f D .r0 ; r1 ; r2 ; : : :/ then f can be written as f D

1 X i D0

ri x i D

m X

ri x i

i D0

for some m  0 since ri ¤ 0 for only finitely many i. Further this presentation is unique. now call x an indeterminate over R and write each element of RQ as f .x/ D PWe m i Q iD0 ri x with f .x/ D 0 or rm ¤ 0. We also now write RŒx for R. Each element of RŒx is called a polynomial over R. The elements r0 ; : : : ; rm are called the coefficients of f .x/ with rm the leading coefficient. If rm ¤ 0 the natural number m is called the degree of f .x/ which we denote by deg f .x/. We say that f .x/ D 0 has degree 1. The uniqueness of the representation of a polynomial implies that two nonzero

54

Chapter 4 Polynomials and Polynomial Rings

polynomials are equal if and only if they have the same degree and exactly the same coefficients. A polynomial of degree 1 is called a linear polynomial while one of degree two is a quadratic polynomial. The set of polynomials of degree 0 together with 0 form a ring isomorphic to R and hence can be identified with R, the constant polynomials. Thus the ring R embeds in the set of polynomials RŒx. The following results are straightforward concerning degree. Lemma 4.1.1. Let f .x/ ¤ 0; g.x/ ¤ 0 2 RŒx. Then: (a) deg f .x/g.x/  deg f .x/ C deg g.x/. (b) deg.f .x/ ˙ g.x//  Max.deg f .x/; deg g.x//. If R is an integral domain then we have equality in (a). Theorem 4.1.2. Let R be a commutative ring with an identity. Then the set of polynomials RŒx forms a ring called the ring of polynomials over R. The ring R identified with 0 and the polynomials of degree 0 naturally embeds into RŒx. RŒx is commutative if and only if R is commutative. Further RŒx is uniquely determined by R and x. P P Proof. Set f .x/ D niD0 ri x i and g.x/ D jmD0 sj x j . The ring properties follow directly by computation. The identification of r 2 R with the polynomial r.x/ D r provides the embedding of R into RŒx. From the definition of multiplication in RŒx if R is commutative then RŒx is commutative. Conversely if RŒx is commutative then from the embedding of R into RŒx it follows that R must be commutative. Note that if R has a multiplicative identity 1 ¤ 0 then this is also the multiplicative identity of RŒx. Finally if S is a ring that contains R and ˛ 2 S then ³ ²X i ri ˛ W ri 2 R and ri ¤ 0 for only a finite number of i RŒ˛ D i 0

is a homomorphic image of RŒx via the map X X ri x i 7! ri ˛ i : i 0

i 0

Hence RŒx is uniquely determined by R and x. If R is an integral domain then irreducible polynomials are defined as irreducibles in the ring RŒx. If R is a field then f .x/ is an irreducible polynomial if there is no factorization f .x/ D g.x/h.x/ where g.x/ and h.x/ are polynomials of lower degree than f .x/. Otherwise f .x/ is called reducible. In elementary mathematics polynomials are considered as functions. We recover that idea via the concept of evaluation.

Section 4.2 Polynomial Rings over Fields

55

Definition 4.1.3. Let f .x/ D r0 C r1 x C    C rm x n be a polynomial over a commutative ring R with an identity and let c 2 R. Then the element f .c/ D r0 C r1 c C    C rn c n 2 R is called the evaluation of f .x/ at c. Definition 4.1.4. If f .x/ 2 RŒx and f .c/ D 0 for c 2 R, then c is called a zero or a root of f .x/ in R.

4.2

Polynomial Rings over Fields

We now restate some of the result of the last chapter for KŒx where K is a field. We then consider some consequences of these results to zeros of polynomials. Theorem 4.2.1. If F is a field, then F Œx forms an integral domain. F can be naturally embedded into F Œx by identifying each element of F with the corresponding constant polynomial. The only units in F Œx are the nonzero elements of F . Proof. Verification of the basic ring properties is solely computational and is left to the exercises. Since deg P .x/Q.x/ D deg P .x/ C deg Q.x/, it follows that if neither P .x/ ¤ 0 nor Q.x/ ¤ 0 then P .x/Q.x/ ¤ 0 and therefore F Œx is an integral domain. If G.x/ is a unit in F Œx, then there exists an H.x/ 2 F Œx with G.x/H.x/ D 1. From the degrees we have deg G.x/ C deg H.x/ D 0 and since deg G.x/  0, deg H.x/  0. This is possible only if deg G.x/ D deg H.x/ D 0. Therefore G.x/ 2 F . Now that we have F Œx as an integral domain we proceed to show that F Œx is a principal ideal domain and hence there is unique factorization into primes. We first repeat the definition of a prime in F Œx. If 0 ¤ f .x/ has no nontrivial, nonunit factors (it cannot be factorized into polynomials of lower degree) then f .x/ is a prime in F Œx or a prime polynomial. A prime polynomial is also called an irreducible polynomial over F . Clearly, if deg g.x/ D 1 then g.x/ is irreducible. The fact that F Œx is a principal ideal domain follows from the division algorithm for polynomials, which is entirely analogous to the division algorithm for integers. Theorem 4.2.2 (division algorithm in F Œx). If 0 ¤ f .x/; 0 ¤ g.x/ 2 F Œx then there exist unique polynomials q.x/; r.x/ 2 F Œx such that f .x/ D q.x/g.x/ C r.x/ where r.x/ D 0 or deg r.x/ < deg g.x/. (The polynomials q.x/ and r.x/ are called respectively the quotient and remainder.) This theorem is essentially long division of polynomials. A formal proof is based on induction on the degree of g.x/. We omit this but give some examples from QŒx.

56

Chapter 4 Polynomials and Polynomial Rings

Example 4.2.3.

(a) Let f .x/ D 3x 4  6x 2 C 8x  6, g.x/ D 2x 2 C 4. Then

3x 4  6x 2 C 8x  6 3 D x 2  6 with remainder 8x C 18: 2 2x C 4 2 Thus here q.x/ D 32 x 2  6 and r.x/ D 8x C 18. (b) Let f .x/ D 2x 5 C 2x 4 C 6x 3 C 10x 2 C 4x, g.x/ D x 2 C x. Then 2x 5 C 2x 4 C 6x 3 C 10x 2 C 4x D 2x 3 C 6x C 4: x2 C x Thus here q.x/ D 2x 3 C 6x C 4 and r.x/ D 0. Theorem 4.2.4. Let F be a field. Then the polynomial ring F Œx is a principal ideal domain and hence a unique factorization domain. We now give some consequences relative to zeros of polynomials in F Œx. Theorem 4.2.5. If f .x/ 2 F Œx and c 2 F with f .c/ D 0 then f .x/ D .x  c/h.x/; where deg h.x/ < deg f .x/. Proof. Divide f .x/ by x  c. Then by the division algorithm we have f .x/ D .x  c/h.x/ C r.x/ where r.x/ D 0 or deg r.x/ < deg.x  c/ D 1. Hence if r.x/ ¤ 0 then r.x/ is a polynomial of degree 0, that is a constant polynomial, that is r.x/ D r for r 2 F . Hence we have f .x/ D .x  c/h.x/ C r: This implies that 0 D f .x/ D 0h.c/ C r D r and therefore r D 0 and f .x/ D .x  c/h.x/. Since deg.x  c/ D 1 we must have that deg h.x/ < deg f .x/. If f .x/ D .x  c/k h.x/ for some k  1 with h.c/ ¤ 0 then c is called a zero of order k. Theorem 4.2.6. Let f .x/ 2 F Œx with degree 2 or 3. Then f is irreducible if and only if f .x/ doesn’t have a zero in F . Proof. Suppose that f .x/ is irreducible of degree 2 or 3. If f .x/ has a zero c then from Theorem 4.2.4 we have f .x/ D .x  c/h.x/ with h.x/ of degree 1 or 2. Therefore f .x/ is reducible a contradiction and hence f .x/ cannot have a zero. Conversely from Theorem 4.2.4 if f .x/ has a zero and if of degree greater than 1 then f .x/ is reducible.

Section 4.3 Polynomial Rings over Integral Domains

4.3

57

Polynomial Rings over Integral Domains

Here we consider RŒx where R is an integral domain. Definition 4.3.1. Let R be an integral domain. Then a1 ; a2 ; : : : ; an 2 R are coprime if the set of all common divisors of a1 ; : : : ; an consists only of units. Notice for example that this concept depends on the ring R. For example 6 and 9 are not coprime over the integers Z since 3j6 and 3j9 and 3 is not a unit. However 6 and 9 are coprime over the rationals Q. Here 3 is a unit. Pn i Definition 4.3.2. Let f .x/ D i D1 ri x 2 RŒx where R is an integral domain. Then f .x/ is a primitive polynomial or just primitive if r0 ; r1 ; : : : ; rn are coprime in R. Theorem 4.3.3. Let R be an integral domain. Then (a) The units of RŒx are the units of R. (b) If p is a prime element of R then p is a prime element of RŒx. Proof. If r 2 R is a unit then since R embeds into RŒx it follows that r is also a unit in RŒx. Conversely suppose that h.x/ 2 RŒx is a unit. Then there is a g.x/ such that h.x/g.x/ D 1. Hence deg f .x/ C deg g.x/ D deg 1 D 0. Since degrees are nonnegative integers it follows that deg f .x/ D deg g.x/ D 0 and hence f .x/ 2 R. Now suppose that p is a prime element of R. Then p ¤ 0 and pR is a prime ideal in R. We must show that pRŒx is a prime ideal in RŒx. Consider the map  W RŒx ! .R=pR/Œx given by  X X n n i ri x D .ri C pR/x i :  i D0

i D0

Then  is an epimorphism with kernel pRŒx. Since pR is a prime ideal we know that R=pR is an integral domain. It follows that .R=pR/Œx is also an integral domain. Hence pRŒx must be a prime ideal in RŒx and therefore p is also a prime element of RŒx. Recall that each integral domain R can be embedded into a unique field of fractions K. We can use results on KŒx to deduce some results in RŒx. Lemma 4.3.4. If K is a field then each nonzero f .x/ 2 KŒx is a primitive. Proof. Since K is a field each nonzero element of K is a unit. Therefore the only common divisors of the coefficients of f .x/ are units and hence f .x/ 2 KŒx is primitive.

58

Chapter 4 Polynomials and Polynomial Rings

Theorem 4.3.5. Let R be an integral domain. Then each irreducible f .x/ 2 RŒx of degree > 0 is primitive. Proof. Let f .x/ be an irreducible polynomial in RŒx and let r 2 R be a common divisor of the coefficients of f .x/. Then f .x/ D rg.x/ where g.x/ 2 RŒx. Then deg f .x/ D deg g.x/ > 0 so g.x/ … R. Since the units of RŒx are the units of R it follows that g.x/ is not a unit in RŒx. Since f .x/ is irreducible it follows that r must be a unit in RŒx and hence r is a unit in R. Therefore f .x/ is primitive. Theorem 4.3.6. Let R be an integral domain and K its field of fractions. If f .x/ 2 RŒx is primitive and irreducible in KŒx then f .x/ is irreducible in RŒx. Proof. Suppose that f .x/ 2 RŒx is primitive and irreducible in KŒx and suppose that f .x/ D g.x/h.x/ where g.x/; h.x/ 2 RŒx  KŒx. Since f .x/ is irreducible in KŒx either g.x/ or h.x/ must be a unit in KŒx. Without loss of generality suppose that g.x/ is a unit in KŒx. Then g.x/ D g 2 K. But g.x/ 2 RŒx and K \RŒx D R. Hence g 2 R. Then g is a divisor of the coefficients of f .x/ and as f .x/ is primitive g.x/ must be a unit in R and therefore also a unit in RŒx. Therefore f .x/ is irreducible in RŒx.

4.4

Polynomial Rings over Unique Factorization Domains

In this section we prove that if R is a UFD then the polynomial ring RŒx is also a UFD. We first need the following due to Gauss. Theorem 4.4.1 (Gauss’ lemma). Let R be a UFD and f .x/; g.x/ primitive polynomials in RŒx. Then their product f .x/g.x/ is also primitive. Proof. Let R be a UFD and f .x/; g.x/ primitive polynomials in RŒx. Suppose that f .x/g.x/ is not primitive. Then there is a prime element p 2 R that divides each of the coefficients of f .x/g.x/. Then pjf .x/g.x/. Since prime elements of R are also prime elements of RŒx it follows that p is also a prime element of RŒx and hence pjf .x/ or pjg.x/. Therefore either f .x/ or g.x/ is not primitive giving a contradiction. Theorem 4.4.2. Let R be a UFD and K its field of fractions. (a) If g.x/ 2 KŒx is nonzero then there is a nonzero a 2 K such that ag.x/ 2 RŒx is primitive. (b) Let f .x/; g.x/ 2 RŒx with g.x/ primitive and f .x/ D ag.x/ for some a 2 K. Then a 2 R. (c) If f .x/ 2 RŒx is nonzero then there is a b 2 R and a primitive g.x/ 2 RŒx such that f .x/ D bg.x/.

Section 4.4 Polynomial Rings over Unique Factorization Domains

59

Pn ri i Proof. (a) Suppose that g.x/ D i D0 ai x with ai D si , ri ; si 2 R. Set s D s0 s1    sn . Then sg.x/ is a nonzero element of RŒx. Let d be a greatest common divisor of the coefficients of sg.x/. If we set a D ds then ag.x/ is primitive. (b) For a 2 K there are coprime r; s 2 R satisfying a D rs . Suppose that a … R. Then there is a prime element p 2 R dividing s. Since g.x/ is primitive p does not divide all the coefficients of g.x/. However we also have f .x/ D ag.x/ D rs g.x/. Hence sf .x/ D rg.x/ where pjs and p doesn’t divide r. Therefore p divides all the coefficients of g.x/ and hence a 2 R. (c) From part (a) there is a nonzero a 2 K such that af .x/ is primitive in RŒx. Then f .x/ D a1 .af .x//. From part (b) we must have a1 2 R. Set g.x/ D af .x/ and b D a1 . Theorem 4.4.3. Let R be a UFD and K its field of fractions. Let f .x/ 2 RŒx be a polynomial of degree  1. (a) If f .x/ is primitive and f .x/jg.x/ in KŒx then f .x/ divides g.x/ also in RŒx. (b) If f .x/ is irreducible in RŒx then it is also irreducible in KŒx. (c) If f .x/ is primitive and a prime element of KŒx then f .x/ is also a prime element of RŒx. Proof. (a) Suppose that g.x/ D f .x/h.x/ with h.x/ 2 KŒx. From Theorem 4.4.2 part (a) there is a nonzero a 2 K such that h1 .x/ D ah.x/ is primitive in RŒx. Hence g.x/ D a1 .f .x/h1 .x/. From Gauss’ lemma f .x/h1 .x/ is primitive in RŒx and therefore from Theorem 4.4.2 part (b) we have a1 2 R. It follows that f .x/jg.x/ in RŒx. (b) Suppose that g.x/ 2 KŒx is a factor of f .x/. From Theorem 4.4.2 part (a) there is a nonzero a 2 K with g1 .x/ D ag.x/ primitive in RŒx. Since a is a unit in K it follows that g.x/jf .x/ in KŒx

implies

g1 .x/jf .x/ in KŒx

and hence since g1 .x/ is primitive g1 .x/jf .x/ in RŒx: However by assumption f .x/ is irreducible in RŒx. This implies that either g1 .x/ is a unit in R or g1 .x/ is an associate of f .x/. If g1 .x/ is a unit then g1 2 K and g1 D ga and hence g 2 K, that is g D g.x/ is a unit. If g1 .x/ is an associate of f .x/ then f .x/ D bg.x/ where b 2 K since g1 .x/ D ag.x/ with a 2 K. Combining these it follows that f .x/ has only trivial factors in KŒx and since by assumption f .x/ is nonconstant it follows that f .x/ is irreducible in KŒx.

60

Chapter 4 Polynomials and Polynomial Rings

(c) Suppose that f .x/jg.x/h.x/ with g.x/; h.x/ 2 RŒx. Since f .x/ is a prime element in KŒx we have that f .x/jg.x/ or f .x/jh.x/ in KŒx. From part (a) we have f .x/jg.x/ or f .x/jh.x/ in RŒx implying that f .x/ is a prime element in RŒx. We can now state and prove our main result. Theorem 4.4.4 (Gauss). Let R be a UFD. Then the polynomial ring RŒx is also a UFD Proof. By induction on degree we show that each nonunit f .x/ 2 RŒx; f .x/ ¤ 0, is a product of prime elements. Since R is an integral domain so is RŒx, and so the fact that RŒx is a UFD then follows from Theorem 3.3.3. If deg f .x/ D 0 then f .x/ D f is a nonunit in R. Since R is a UFD f is a product of prime elements in R. However from Theorem 4.3.3 each prime factor is then also prime in RŒx. Therefore f .x/ is a product of prime elements. Now suppose n > 0 and that the claim is true for all polynomials f .x/ of degree < n. Let f .x/ be a polynomial of degree n > 0. From Theorem 4.4.2 (c) there is an a 2 R and a primitive h.x/ 2 RŒx satisfying f .x/ D ah.x/. Since R is a UFD the element a is a product of prime elements in R or a is a unit in R. Since the units in RŒx are the units in R and a prime element in R is also a prime element in RŒx it follows that a is a product of prime elements in RŒx or a is a unit in RŒx. Let K be the field of fractions of R. Then KŒx is a UFD. Hence h.x/ is a product of prime elements of KŒx. Let p.x/ 2 KŒx be a prime divisor of h.x/. From Theorem 4.4.2 we can assume by multiplication of field elements that p.x/ 2 RŒx and p.x/ is primitive. From Theorem 4.4.2 (c) it follows that p.x/ is a prime element of RŒx and further from Theorem 4.4.3 (a) that p.x/ is a divisor of h.x/ in RŒx. Therefore f .x/ D ah.x/ D ap.x/g.x/ 2 RŒx where (1) a is a product of prime elements of RŒx or a is a unit in RŒx, (2) deg p.x/ > 0, since p.x/ is a prime element in KŒx, (3) p.x/ is a prime element in RŒx, and (4) deg g.x/ < deg f .x/ since deg p.x/ > 0. By our inductive hypothesis we have then that g.x/ is a product of prime elements in RŒx or g.x/ is a unit in RŒx. Therefore the claim holds for f .x/ and therefore holds for all f .x/ by induction. If RŒx is a polynomial ring over R we can form a polynomial ring in a new indeterminate y over this ring to form .RŒx/Œy. It is straightforward that .RŒx/Œy is isomorphic to .RŒy/Œx. We denote both of these rings by RŒx; y and consider this as the ring of polynomials in two commuting variables x; y with coefficients in R.

Section 4.4 Polynomial Rings over Unique Factorization Domains

61

If R is a UFD then from Theorem 4.4.4 RŒx is also a UFD and hence RŒx; y is also a UFD. Inductively then the ring of polynomials in n commuting variables RŒx1 ; x2 ; : : : ; xn  is also a UFD. Corollary 4.4.5. If R is a UFD then the polynomial ring in n commuting variables RŒx1 ; : : : ; xn  is also a UFD. We now give a condition for a polynomial in RŒx to have a zero in KŒx where K is the field of fractions of R. Theorem 4.4.6. Let R be a UFD and K its field of fractions. Let f .x/ D x n C rn1 x n1 C    C r0 2 RŒx. Suppose that ˇ 2 K is a zero of f .x/. Then ˇ is in R and is a divisor of r0 . Proof. Let ˇ D

r s

where s ¤ 0 and r; s 2 R and r; s are coprime. Now f

  r n1 rn r D 0 D n C rn1 n1 C    C r0 : s s s

Hence it follows that s must divide r n . Since r and s are coprime s must be a unit and then without loss of generality we may assume that s D 1. Then ˇ 2 R and r.r n1 C    C a1 / D a0 and so rja0 . Note that since Z is a UFD, Gauss’ theorem implies that ZŒx is also a UFD. However ZŒx is not a principal ideal domain. For example the set of integral polynomials with even constant term is an ideal but not principal. We leave the verification to the exercises. On the other hand we saw that if K is a field KŒx is a PID. The question arises as to when RŒx actually is a principal ideal domain. It turns out to be precisely when R is a field. Theorem 4.4.7. Let R be a commutative ring with an identity. Then the following are equivalent: (1) R is a field. (2) RŒx is Euclidean. (3) RŒx is a principal ideal domain. Proof. From Section 4.2 we know that (a) implies (b) which in turn implies (c). Therefore we must show that (c) implies (a). Assume then that RŒx is a principal ideal domain. Define the map  W RŒx ! R

62

Chapter 4 Polynomials and Polynomial Rings

by  .f .x// D f .0/: It is easy to see that  is a ring homomorphism with RŒx= ker. / Š R. Therefore ker. / ¤ RŒx. Since RŒx is a principal ideal domain it is an integral domain. It follows that ker. / must be a prime ideal since the quotient ring is an integral domain. However since RŒx is a principal ideal domain prime ideals are maximal ideals and hence ker. / is a maximal ideal. Therefore R Š RŒx= ker. / is a field. We now consider the relationship between irreducibles in RŒx for a general integral domain and irreducibles in KŒx where K is its field of fractions. This is handled by the next result called Eisenstein’s criterion. Theorem 4.4.8 (Eisenstein’s Let R be an integral domain and K its field Pcriterion). n i 2 RŒx of degree n > 0. Let p be a prime a x of fractions. Let f .x/ D i D0 i element of R satisfying (1) pjai for i D 0; : : : ; n  1. (2) p does not divide an . (3) p 2 does not divide a0 . Then: (a) If f .x/ is primitive then f .x/ is irreducible in RŒx. (b) Suppose that R is a UFD. Then f .x/ is also irreducible in KŒx. Proof. (a) Suppose that f .x/ D g.x/h.x/ with g.x/; h.x/ 2 RŒx. Suppose that g.x/ D

k X

bi x i ;

bk ¤ 0 and

h.x/ D

i D1

l X

cj x j ;

cl ¤ 0:

i D1

Then a0 D b0 c0 . Now pja0 but p 2 does not divide a0 . This implies that either p doesn’t divide b0 or p doesn’t divide c0 . Without loss of generality assume that pjb0 and p doesn’t divide c0 . Since an D bk cl and p does not divide an it follows that p does not divide bk . Let bj be the first coefficient of g.x/ which is not divisible by p. Consider aj D bj c0 C    C b0 cj where everything after the first term is divisible by p. Since p does not divide both bj and c0 it follows that p does not divide bj c0 and therefore p does not divide aj which implies that j D n. Then from j  k  n it follows that k D n. Therefore deg g.x/ D deg f .x/ and hence deg h.x/ D 0. Thus h 2 R. Then from f .x/ D hg.x/ with f primitive it follows that h is a unit and therefore f .x/ is irreducible.

Section 4.4 Polynomial Rings over Unique Factorization Domains

63

(b) Suppose that f .x/ D g.x/h.x/ with g.x/; h.x/ 2 RŒx. The fact that f .x/ was primitive was only used in the final part of part (a) so by the same arguments as in part (a) we may assume without loss of generality that h 2 R  K. Therefore f .x/ is irreducible in KŒx. We give some examples. Example 4.4.9. Let R D Z and p a prime number. Suppose that n; m are integers such that n  1 and p does not divide m. Then x n ˙ pm is irreducible in ZŒx and 1 QŒx. In particular .pm/ n is irrational. Example 4.4.10. Let R D Z and p a prime number. Consider the polynomial ˆp .x/ D

xp  1 D x p1 D x p2 C    C 1: x1

Since all the coefficients of ˆp .x/ are equal to 1, Eisenstein’s criterion is not directly applicable. However the fact that ˆp .x/ is irreducible implies that for any integer a the polynomial ˆp .x C a/ is also irreducible in ZŒx. It follows that    p  x p C p1 x p1 C    C p1 x C 1p  1 .x C 1/p  1 D ˆp .x C 1/ D .x C 1/  1 x ! ! p p1 p D x p1 C x C  C : 1 p1  p    D p is not Now pj pi for 1  i  p  1 (see exercises) and moreover p1 2 divisible by p . Therefore we can apply the Eisenstein criterion to conclude that ˆp .x/ is irreducible in ZŒx and QŒx. P Theorem 4.4.11. Let R be a UFD and K its field of fractions. Let f .x/ D niD0 ai x i 2 RŒx be a polynomial of degree  1. Let P be a prime ideal in R with an … P . Let R D R=P and let ˛ W RŒx ! RŒx be defined by ˛

X m i D0

 ri x

i

D

m X

.ri C P /x i :

i D0

˛ is an epimorphism. Then if ˛.f .x// is irreducible in RŒx then f .x/ is irreducible in KŒx. Proof. By Theorem 4.4.3 there is an a 2 R and a primitive g.x/ 2 RŒx satisfying f .x/ D ag.x/. Since an … P we have that ˛.a/ ¤ 0 and further the highest coefficient of g.x/ is also not an element of P . If ˛.g.x// is reducible then ˛.f .x// is also reducible. Thus ˛.g.x// is irreducible. However from Theorem 4.4.4 g.x/ is

64

Chapter 4 Polynomials and Polynomial Rings

irreducible in KŒx so f .x/ D ag.x/ is also irreducible in KŒx. Therefore to prove the theorem it suffices to consider the case where f .x/ is primitive in RŒx. Now suppose that f .x/ is primitive. We show that f .x/ is irreducible in RŒx. Suppose that f .x/ D g.x/h.x/, g.x/; h.x/ 2 RŒx with f .x/; g.x/ nonunits in RŒx. Since f .x/ is primitive g; h … R and so deg g.x/ < deg f .x/ and deg h.x/ < deg f .x/. Now we have ˛.f .x// D ˛.g.x//˛.h.x//. Since P is a prime ideal R=P is an integral domain and so in RŒx we have deg ˛.g.x// C deg ˛.g.x// D deg ˛.f .x// D deg f .x/ since an … P . Since R is a UFD it has no zero divisors and so deg f .x/ D deg g.x/ C deg h.x/: Now deg ˛.g.x//  deg g.x/ deg ˛.h.x//  deg h.x/: Therefore deg ˛.g.x// D deg g.x/ and deg ˛.h.x// D deg h.x/. Therefore ˛.f .x// is reducible and we have a contradiction. It is important to note that ˛.f .x// being reducible does not imply that f .x/ is reducible. For example f .x/ D x 2 C 1 is irreducible in ZŒx. However in Z2 Œx we have x 2 C 1 D .x C 1/2 and hence f .x/ is reducible in Z2 Œx. Example 4.4.12. Let f .x/ D x 5  x 2 C 1 2 ZŒx. Choose P D 2Z so that ˛.f .x// D x 5 C x 2 C 1 2 Z2 Œx: Suppose that in Z2 Œx we have ˛.f .x// D g.x/h.x/. Without loss of generality we may assume that g.x/ is of degree 1 or 2. If deg g.x/ D 1 then ˛.f .x// has a zero c in Z2 Œx. The two possibilities for c are c D 0 or c D 1. Then If c D 0 then 0 C 0 C 1 D 1 ¤ 0: If c D 1 then 1 C 1 C 1 D 1 ¤ 0: Hence the degree of g.x/ cannot be 1.

65

Section 4.5 Exercises

Suppose deg g.x/ D 2. The polynomials of degree 2 over Z2 Œx have the form x 2 C x C 1;

x 2 C x;

x 2 C 1;

x2:

The last three, x 2 Cx; x 2 C1; x 2 all have zeros in Z2 Œx so they can’t divide ˛.f .x//. Therefore g.x/ must be x 2 C x C 1. Applying the division algorithm we obtain ˛.f .x// D .x 3 C x 2 /.x 2 C x C 1/ C 1 and therefore x 2 CxC1 does not divide ˛.f .x//. It follows that ˛.f .x// is irreducible and from the previous theorem f .x/ must be irreducible in QŒx.

4.5

Exercises

1. For which a; b 2 Z does the polynomial x 2 C 3x C 1 divide the polynomial x 3 C x 2 C ax C b? 2. Let a C bi 2 C be a zero of f .x/ 2 RŒx. Show that also a  i b is a zero of f .x/. 3. Determine all irreducible polynomials over R. Factorize f .x/ 2 RŒx in irreducible polynomials. 4. Let R be an integral domain, A G R an ideal and f 2 RŒx a monic P polynomial. N Define f 2 RŒx=AŒx by the mapping RŒx ! R=AŒx, f D ai x i 7! fN D P i N aNi x , where aN WD a C A. Show: If f 2 R=AŒx is irreducible then also f 2 RŒx is irreducible. 5. Decide if the following polynomials f 2 RŒx are irreducible: (i) f .x/ D x 3 C 2x 2 C 3, R D Z. (ii) f .x/ D x 5  2x C 1, R D Q. (iii) f .x/ D 3x 4 C 7x 2 C 14x C 7, R D Q. (iv) f .x/ D x 7 C .3  i /x 2 C .3 C 4i /x C 4 C 2i , R D ZŒi. (v) f .x/ D x 4 C 3x 3 C 2x 2 C 3x C 4, R D Q. (vi) f .x/ D 8x 3  4x 2 C 2x  1, R D Z. 6. Let R be an integral domain with characteristic 0, let k  1 and ˛ 2 R. In RŒx define the derivatives f .k/ .x/, k D 0; 1; 2; : : : , of a polynomial f .x/ 2 RŒx by f 0 .x/ WD f .x/; 0

f .k/ .x/ WD f .k1/ .x/: Show that ˛ is a zero of order k of the polynomial f .x/ 2 RŒx, if f .k1/ .˛/ D 0, but f .k/ .˛/ ¤ 0. 7. Prove that the set of integral polynomials with even constant term is an ideal but not principal.   8. Prove that pj pi for 1  i  p  1.

Chapter 5

Field Extensions

5.1

Extension Fields and Finite Extensions

Much of algebra in general arose from the theory of equations, specifically polynomial equations. As discovered by Galois and Abel the solutions of polynomial equations over fields is intimately tied to the theory of field extensions. This theory eventually blossoms into Galois Theory. In this chapter we discuss the basic material concerning field extensions. Recall that if L is a field and K  L is also a field under the same operations as L then K is called a subfield of L. If we view this situation from the viewpoint of K we say that L is an extension field or field extension of K. If K; L are fields with K  L we always assume that K is a subfield of L. Definition 5.1.1. If K; L are fields with K  L then we say that L is a field extension or extension field of K. We denote this by LjK. Note that this is equivalent to having a field monomorphism i WK!L and then identifying K and i.K/. As examples we have that R is an extension field of Q and C is an extension field of both C and Q. If K is any field then the ring of polynomials KŒx over K is an integral domain. Let K.x/ be the field of fractions of KŒx. This is called the field of rational functions over K. Since K can be considered as part of KŒx it follows that K  K.x/ and hence K.x/ is an extension field of K. A crucial concept is that of the degree of a field extension. Recall that a vector space V over a field F consists of an abelian group V together with scalar multiplication from F satisfying: (1) f v 2 V if f 2 F , v 2 V . (2) f .u C v/ D f u C f v for f 2 F , u; v 2 V . (3) .f C g/v D f v C gv for f; g 2 F , v 2 V . (4) .fg/v D f .gv/ for f; g 2 F , v 2 V . (5) 1v D v for v 2 V .

67

Section 5.1 Extension Fields and Finite Extensions

Notice that if K is a subfield of L then multiplication of elements of L by elements of K are still in L. Since L is an abelian group under addition, L can be considered as a vector space over K. Thus any extension field is a vector space over any of its subfields. Using this we define the degree jL W Kj of an extension K  L as the dimension dimK .L/ of L as a vector space over K. We call L a finite extension of K if jL W Kj < 1. Definition 5.1.2. If L is an extension field of K then the degree of the extension LjK is defined as the dimension, dimK .L/, of L, as a vector space over K. We denote the degree by jL W Kj. The field extension LjK is a finite extension if the degree jL W Kj is finite. Lemma 5.1.3. jC W Rj D 2 but jR W Qj D 1. Proof. Every complex number can be written uniquely as a C i b where a; b 2 R. Hence the elements 1; i constitute a basis for C over R and therefore the dimension is 2, that is jC W Rj D 2. The fact that jR W Qj D 1 depends on the existence of transcendental numbers. An element r 2 R is algebraic (over Q) if it satisfies some nonzero polynomial with coefficients from Q. That is, P .r/ D 0, where 0 ¤ P .x/ D a0 C a1 x C    C an x n

with ai 2 Q:

Any q 2 Q is algebraic since if P .x/ D p x  q then P .q/ D 0. However, many p irrationals are also algebraic. For example 2 is algebraic since x 2  2 D 0 has 2 as a root. An element r 2 R is transcendental if it is not algebraic. In general it is very difficult to show that a particular element is transcendental. However there are uncountably many transcendental elements (see exercises). Specific examples are e and . We will give a proof of their transcendence later in this book. Since e is transcendental, for any natural number n the set of vectors ¹1; e; e 2 ; : : : ; e n º must be independent over Q, for otherwise there would be a polynomial that e would satisfy. Therefore, we have infinitely many independent vectors in R over Q which would be impossible if R had finite degree over Q. Lemma 5.1.4. If K is any field then jK.x/ W Kj D 1. Proof. For any n the elements 1; x; x 2 ; : : : ; x n are independent over K. Therefore as in the proof of Lemma 5.1.3 K.x/ must be infinite dimensional over K. If LjK and L1 jK1 are field extensions then they are isomorphic field extensions if there exists a field isomorphism f W L ! L1 such that fjK is an isomorphism from K to K1 . Suppose that K  L  M are fields. Below we show that the degrees multiply. In this situation where K  L  M we call L an intermediate field.

68

Chapter 5 Field Extensions

Theorem 5.1.5. Let K; L; M be fields with K  L  M . Then jM W Kj D jM W LjjL W Kj: Note that jM W Kj D 1 if and only if either jM W Lj D 1 or jL W Kj D 1. Proof. Let ¹xi W i 2 I º be a basis for L as a vector space over K and let ¹yj W j 2 J º be a basis for M as a vector space over L. To prove the result it is sufficient to show that the set B D ¹xi yj W i 2 I; j 2 J º is a basis for M as a vector space over K. To show this we must show that B is a linearly independent set over K and that B spans M . Suppose that X kij xi yj D 0 where kij 2 K: i;j

We can then write this sum as XX j

 kij xi yj D 0:

i

P But i kij xi 2 L. Since ¹yj W j 2 JP º is a basis for M over L the yj are independent over L and hence for each j we get, i kij xi D 0. Now since ¹xi W i 2 I º is a basis for LP over K it follows that the xi are linearly independent and since for each j we have i kij xi D 0 it must be that kij D 0 for all i and for all j . Therefore the set B is linearly independent over K. Now suppose that m 2 M . Then since ¹yj W j 2 J º spans M over L we have mD

X

cj yj

with cj 2 L:

j

However ¹xi W i 2 I º spans L over K and so for each cj we have cj D

X

kij xi

with kij 2 K:

i

Combining these two sums we have mD

X

kij xi yj

ij

and hence B spans M over K. Therefore B is a basis for M over K and the result is proved.

69

Section 5.2 Finite and Algebraic Extensions

Corollary 5.1.6. (a) If jL W Kj is a prime number then there exists no proper intermediate field between L and K. (b) If K  L and jL W Kj D 1 then L D K. Let LjK be a field extension and suppose that A  L. Then certainly there are subrings of L containing both A and K, for example L. We denote by KŒA the intersection of all subrings of L containing both K and A. Since the intersection of subrings is a subring it follows that KŒA is a subring containing both K and A and the smallest such subring. We call KŒA the ring adjunction of A to K. In an analogous manner we let K.A/ be the intersection of all subfields of L containing both K and A. This is then a subfield of L and the smallest subfield of L containing both K and A. The subfield K.A/ is called the field adjunction of A to K. Clearly KŒA  K.A/. If A D ¹a1 ; : : : ; an º then we write KŒA D KŒa1 ; : : : ; an  and

K.A/ D K.a1 ; : : : ; an /:

Definition 5.1.7. The field extension LjK is finitely generated if there exist a1 ; : : : ; an 2 L such that K D K.a1 ; : : : ; an /. The extension LjK is a simple extension if there is an a 2 L with L D K.a/. In this case a is called a primitive element of LjK. Later we will look at an alternative way to view the adjunction constructions in terms of polynomials.

5.2

Finite and Algebraic Extensions

We now turn to the relationship between field extensions and the solution of polynomial equations. Definition 5.2.1. Let LjK be a field extension. An element a 2 L is algebraic over K if there exists a polynomial p.x/ 2 KŒx with p.a/ D 0. L is an algebraic extension of K if each element of L is algebraic over K. An element a 2 L that is not algebraic over K is called transcendental. L is a transcendental extension if there are transcendental elements, that is they are not algebraic over K. For the remainder of this section we assume that LjK is a field extension. Lemma 5.2.2. Each element of K is algebraic over K. Proof. Let k 2 K. Then k is a root of the polynomial p.x/ D x  k 2 KŒx.

70

Chapter 5 Field Extensions

We tie now algebraic extensions to finite extensions. Theorem 5.2.3. If LjK is a finite extension then LjK is an algebraic extension. Proof. Suppose that LjK is a finite extension and a 2 L. We must show that a is algebraic over K. Suppose that jL W Kj D n < 1 then dimK .L/ D n. It follows that any n C 1 elements of L are linearly dependent over K. Now consider the elements 1; a; a 2 ; : : : ; an in L. These are n C 1 distinct elements in L so they are dependent over K. Hence there exist c0 ; : : : ; cn 2 K not all zero such that c0 C c1 a C    C cn an D 0: Let p.x/ D c0 C c1 x C    C cn x n . Then p.x/ 2 KŒx and p.a/ D 0. Therefore a is algebraic over K. Since a was arbitrary it follows that L is an algebraic extension of K. From the previous theorem it follows that every finite extension is algebraic. The converse is not true, that is there are algebraic extensions that are not finite. We will give examples in Section 5.4. The following lemma gives some examples of algebraic and transcendental extensions. Lemma 5.2.4. CjR is algebraic but RjQ and CjQ are transcendental. If K is any field then K.x/jK is transcendental. Proof. Since 1; i constitute a basis for C over R we have jC W Rj D 2. Hence C is a finite extension of R and therefore from Theorem 5.2.3 an algebraic extension. More directly if ˛ D a C i b 2 C then ˛ is a zero of x 2  2ax C .a2 C b 2 / 2 RŒx. The existence of transcendental numbers (we will discuss these more fully in Section 5.5) shows that both RjQ and CjQ are transcendental extensions. Finally the element x 2 K.x/ is not a zero of any polynomial in KŒx. Therefore x is a transcendental element so the extension K.x/jK is transcendental.

5.3

Minimal Polynomials and Simple Extensions

If LjK is a field extension and a 2 L is algebraic over K then p.a/ D 0 for some polynomial p.x/ 2 KŒx. In this section we consider the smallest such polynomial and tie it to a simple extension of K. Definition 5.3.1. Suppose that LjK is a field extension and a 2 L is algebraic over K. The polynomial ma .x/ 2 KŒx is the minimal polynomial of a over K if

Section 5.3 Minimal Polynomials and Simple Extensions

71

(1) ma .x/ has leading coefficient 1, that is, it is a monic polynomial. (2) ma .a/ D 0. (3) If f .x/ 2 KŒx with f .a/ D 0 then ma .x/jf .x/. Hence ma .x/ is the monic polynomial of minimal degree that has a as a zero. We prove next that every algebraic element has such a minimal polynomial. Theorem 5.3.2. Suppose that LjK is a field extension and a 2 L is algebraic over K. Then (1) The minimal polynomial ma .x/ 2 KŒx exists and is irreducible over K. (2) KŒa Š K.a/ Š KŒx=.ma .x// where .ma .x// is the principal ideal in KŒx generated by ma .x/. (3) jK.a/ W Kj D deg.ma .x//. Therefore K.a/jK is a finite extension. Proof. (1) Suppose that a 2 L is algebraic over K. Let I D ¹f .x/ 2 KŒx W f .a/ D 0º: Since a is algebraic I ¤ ;. It is straightforward to show (see exercises) that I is an ideal in KŒx. Since K is a field we have that KŒx is a principal ideal domain. Hence there exists g.x/ 2 KŒx with I D .g.x//. Let b be the leading coefficient of g.x/. Then ma .x/ D b 1 g.x/ is a monic polynomial. We claim that ma .x/ is the minimal polynomial of a and that ma .x/ is irreducible. First it is clear that I D .g.x// D .ma .x//. If f .x/ 2 KŒx with f .a/ D 0 then f .x/ D h.x/ma .x/ for some h.x/. Therefore ma .x/ divides any polynomial that has a as a zero. It follows that ma .x/ is the minimal polynomial. Suppose that ma .x/ D g1 .x/g2 .x/. Then since ma .a/ D 0 it follows that either g1 .a/ D 0 or g2 .a/ D 0. Suppose g1 .a/ D 0. Then from above ma .x/jg1 .x/ and since g1 .x/jma .x/ we must then have that g2 .x/ is a unit. Therefore ma .x/ is irreducible. (2) Consider the map  W KŒx ! KŒa given by X  X i ki x D ki a i :  i

i

Then  is a ring epimorphism (see exercises) and ker. / D ¹f .x/ 2 KŒx W f .a/ D 0º D .ma .x// from the argument in the proof of part (1). It follows that KŒx=.ma .x// Š KŒa:

72

Chapter 5 Field Extensions

Since ma .x/ is irreducible we have KŒx=.ma .x// is a field and therefore KŒa D K.a/. (3) Let n D deg.ma .x//. We claim that the elements 1; a; : : : ; an1 are a basis for KŒa D K.a/ over K. First suppose that n1 X

ci a i D 0

i D1

P i with not all ci D 0 and ci 2 K. Then h.a/ D 0 where h.x/ D n1 i D0 ci x . But this contradicts the fact that ma .x/ has minimal degree over all polynomials in KŒx that have a as a zero. Therefore the set 1; a; : : : ; an1 is linearly independent over K. Now let b 2 KŒa Š KŒx=.ma .x//. Then there is a g.x/ 2 KŒx with b D g.a/. By the division algorithm g.x/ D h.x/ma .x/ C r.x/ where r.x/ D 0 or deg.r.x// < deg.ma .x//. Now r.a/ D g.a/  h.a/ma .a/ D g.a/ D b: If r.x/ D 0 then b D 0. If r.x/ ¤ 0 then since deg.r.x// < n we have r.x/ D c0 C c1 x C    C cn1 x n1 with ci 2 K and some ci but not all might be zero. This implies that b D r.a/ D c0 C c1 a C    C cn1 an1 and hence b is a linear combination over K of 1; a; : : : ; an1 . Hence 1; a; : : : ; an1 spans KŒa over K and hence forms a basis. Theorem 5.3.3. Suppose that LjK is a field extension and a 2 L is algebraic over K. Suppose that f .x/ 2 KŒx is a monic polynomial with f .a/ D 0. Then f .x/ is the minimal polynomial if and only if f .x/ is irreducible in KŒx. Proof. Suppose that f .x/ is the minimal polynomial of a. Then f .x/ is irreducible from the previous theorem. Conversely suppose that f .x/ is monic, irreducible and f .a/ D 0. From the previous theorem ma .x/jf .x/. Since f .x/ is irreducible we have f .x/ D cma .x/ with c 2 K. However since both f .x/ and ma .x/ are monic we must have c D 1 and f .x/ D ma .x/. We now show that a finite extension of K is actually finitely generated over K and further it is generated by finitely many algebraic elements.

Section 5.3 Minimal Polynomials and Simple Extensions

73

Theorem 5.3.4. Let LjK be a field extension. Then the following are equivalent: (1) LjK is a finite extension. (2) LjK is an algebraic extension and there exist elements a1 ; : : : ; an 2 L such that L D K.a1 ; : : : ; an /. (3) There exist algebraic elements a1 ; : : : ; an 2 L such that L D K.a1 ; : : : ; an /. Proof. (1) ) (2). We have seen in Theorem 5.2.3 that a finite extension is algebraic. Suppose that a1 ; : : : ; an are a basis for L over K. Then clearly L D K.a1 ; : : : ; an /. (2) ) (3). If LjK is an algebraic extension and L D K.a1 ; : : : ; an / then each ai is algebraic over K. (3) ) (1). Suppose that there exist algebraic elements a1 ; : : : ; an 2 L such that L D K.a1 ; : : : ; an /. We show that LjK is a finite extension. We do this by induction on n. If n D 1 then L D K.a/ for some algebraic element a and the result follows from Theorem 5.3.2. Suppose now that n  2. We assume then that an extension K.a1 ; : : : ; an1 / with a1 ; : : : ; an1 algebraic elements is a finite extension. Now suppose that we have L D K.a1 ; : : : ; an / with a1 ; : : : ; an algebraic elements. Then jK.a1 ; : : : ; an / W Kj D jK.a1 ; : : : ; an1 /.an / W K.a1 ; : : : ; an1 /jjK.a1 ; : : : ; an1 / W Kj: The second term jK.a1 ; : : : ; an1 / W Kj is finite from the inductive hypothesis. The first term jK.a1 ; : : : ; an1 /.an / W K.a1 ; : : : ; an1 /j is also finite from Theorem 5.3.2 since it is a simple extension of the field K.a1 ; : : : ; an1 / by the algebraic element an . Therefore jK.a1 ; : : : ; an / W Kj is finite. Theorem 5.3.5. Suppose that K is a field and R is an integral domain with K  R. Then R can be viewed as a vector space over K. If dimK .R/ < 1 then R is a field. Proof. Let r0 2 R with r0 ¤ 0. Define the map from R to R given by  .r/ D rr0 : It is easy to show (see exercises) that this is a linear transformation from R to R considered as a vector space over K. Suppose that  .r/ D 0. Then rr0 D 0 and hence r D 0 since r0 ¤ 0 and R is an integral domain. It follows that  is an injective map. Since R is a finite dimensional vector space over K and  is an injective linear transformation it follows that  must also be surjective. This implies that there exists and r1 with  .r1 / D 1. Then r1 r0 D 1 and hence r0 has an inverse within R. Since r0 was an arbitrary nonzero element of R it follows that R is a field.

74

Chapter 5 Field Extensions

Theorem 5.3.6. Suppose that K  L  M is a chain of field extensions. Then M jK is algebraic if and only if M jL is algebraic and LjK is algebraic. Proof. If M jK is algebraic then certainly M jL and LjK are algebraic. Now suppose that M jL and LjK are algebraic. We show that M jK is algebraic. Let a 2 M . Then since a is algebraic over L there exist b0 ; b1 ; : : : ; bn 2 L with b0 C b1 a C    C bn an D 0: Each bi is algebraic over K and hence K.b0 ; : : : ; bn / is finite dimensional over K. Therefore K.b0 ; : : : ; bn /.a/ D K.b0 ; : : : ; bn ; a/ is also finite dimensional over K. Therefore K.b0 ; : : : ; bn ; a/ is a finite extension of K and hence an algebraic extension K. Since a 2 K.b0 ; : : : ; bn ; a/ it follows that a is algebraic over K and therefore M is algebraic over K.

5.4

Algebraic Closures

As before suppose that LjK is a field extension. Since each element of K is algebraic over K there are certainly algebraic elements over K within L. Let AK denote the set of all elements of L that are algebraic over K. We prove that AK is actually a subfield of L. It is called the algebraic closure of K within L. Theorem 5.4.1. Suppose that LjK is a field extension and let AK denote the set of all elements of L that are algebraic over K. Then AK is a subfield of L. AK is called the algebraic closure of K in L. Proof. Since K  AK we have that AK ¤ ;. Let a; b 2 AK . Since a; b are both algebraic over K from Theorem 5.3.4 we have that K.a; b/ is a finite extension of K. Therefore K.a; b/ is an algebraic extension of K and hence each element of K.a; b/ is algebraic over K. Now a; b 2 K.a; b/, if b ¤ 0, and K.a; b/ is a field so a ˙ b; ab and a=b are all in K.a; b/ and hence all algebraic over K. Therefore a ˙ b; ab; a=b, if b ¤ 0, are all in AK . It follows that AK is a subfield of L. In Section 5.2 we showed that every finite extension is an algebraic extension. We mentioned that the converse is not necessarily true, that is there are algebraic extensions that are not finite. Here we give an example. Theorem 5.4.2. Let A be the algebraic closure of the rational numbers Q within the complex numbers C. Then A is an algebraic extension of Q but jA W Qj D 1. Proof. From the previous theorem A is an algebraic extension of Q. We show that it cannot be a finite extension. By Eisenstein’s criterion the rational polynomial f .x/ D x p C p is irreducible over Q for any prime p. Let a be a zero in C of f .x/. Then

Section 5.5 Algebraic and Transcendental Numbers

75

a 2 A and jQ.a/ W Qj D p. Therefore jA W Qj  p for all primes p. Since there are infinitely many primes this implies that jA W Qj D 1

5.5

Algebraic and Transcendental Numbers

In this section we consider the string of field extensions Q  R  C. Definition 5.5.1. An algebraic number ˛ is an element of C which is algebraic over Q. Hence an algebraic number is an ˛ 2 C such that f .˛/ D 0 for some f .x/ 2 QŒx. If ˛ 2 C is not algebraic it is transcendental. We will let A denote the totality of algebraic numbers within the complex numbers C, and T the set of transcendentals so that C D A [ T . In the language of the last subsection, A is the algebraic closure of Q within C. As in the general case, if ˛ 2 C is algebraic we will let m˛ .x/ denote the minimal polynomial of ˛ over Q. We now examine the sets A and T more closely. Since A is precisely the algebraic closure of Q in C we have from our general result that A actually forms a subfield of C. Further since the intersection of subfields is again a subfield it follows that A0 D A \ R the real algebraic numbers form a subfield of the reals. Theorem 5.5.2. The set A of algebraic numbers forms a subfield of C. The subset A0 D A \ R of real algebraic numbers forms a subfield of R. Since each rational is algebraic it is clear p that there are algebraic numbers. Further there are irrational algebraic numbers, 2 for example, since it satisfies the irreducible polynomial x 2  2 D 0 over Q. On the other hand we haven’t examined the question of whether transcendental numbers really exist. To show that any particular complex number is transcendental is in general quite difficult. However it is relatively easy to show that there are uncountably infinitely many transcendentals. Theorem 5.5.3. The set A of algebraic numbers is countably infinite. Therefore T the set of transcendental numbers and T 0 D T \ R, the real transcendental numbers, are uncountably infinite. Proof. Let Pn D ¹f .x/ 2 QŒx W deg.f .x//  nº: Since if f .x/ 2 Pn , f .x/ D qo C q1 x C    C qn x n with qi 2 Q we can identify a polynomial of degree  n with an .n C 1/-tuple .q0 ; q1 ; : : : ; qn / of rational numbers. Therefore the set Pn has the same size as the .n C 1/-fold Cartesian product of Q: QnC1 D Q  Q      Q:

76

Chapter 5 Field Extensions

Since a finite Cartesian product of countable sets is still countable it follows that Pn is a countable set. Now let [ Bn D ¹roots ofp.x/º; p.x/2Pn

that is Bn is the union of all roots in C of all rational polynomials of degree  n. Since each such p.x/ has a maximum of n roots and since Pn is countable it follows that Bn is a countable union of finite sets and hence is still countable. Now AD

1 [

Bn

nD1

so A is a countable union of countable sets and is therefore countable. Since both R and C are uncountably infinite the second assertions follow directly from the countability of A. If say T were countable then C D A [ T would also be countable which is a contradiction. From Theorem 5.5.3 we know that there exist infinitely many transcendental numbers. Liouville in 1851 gave the first proof of the existence of transcendentals by exhibiting a few. He gave as one the following example. Theorem 5.5.4. The real number cD

1 X 1 10j Š j D1

is transcendental. P Proof. First of all since 101j Š < 101j , and j1D1 101j is a convergent geometric series, it follows from the comparison test P that the infinite series defining c converges and defines a real number. Further since j1D1 101j D 19 , it follows that c < 19 < 1. Suppose that c is algebraic so that g.c/ D 0 for some rational nonzero polynomial g.x/. Multiplying through by the least common multiple of all the denominators Pn in g.x/j we may suppose that f .c/ D 0 for some integral polynomial f .x/ D j D0 mj x . Then c satisfies n X j D0

for some integers m0 ; : : : ; mn .

mj c j D 0

77

Section 5.5 Algebraic and Transcendental Numbers

If 0 < x < 1 then by the triangle inequality ˇ n ˇ X n ˇX ˇ 0 j 1 ˇ ˇ j mj x jj mj j D B  jf .x/j D ˇ ˇ j D1

j D1

where B is a real constant depending only on the coefficients of f .x/. Now let k X 1 ck D 10j Š j D1

be the k-th partial sum for c. Then jc  ck j D

1 X j DkC1

1 1 < 2  .kC1/Š : j Š 10 10

Apply the mean value theorem to f .x/ at c and ck to obtain jf .c/  f .ck /j D jc  ck jjf 0 . /j for some with ck < < c < 1. Now since 0 < < 1 we have jc  ck jjf 0 . /j < 2B

1 10.kC1/Š

:

On the other hand, since f .x/ can have at most n roots, it follows that for all k large enough we would have f .ck / ¤ 0. Since f .c/ D 0 we have ˇ ˇX ˇ n 1 j ˇˇ ˇ jf .c/  f .ck /j D jf .ck /j D ˇ mj ck ˇ > nkŠ 10 j D1 since for each j , mj ckj is a rational number with denominator 10j kŠ . However if k is chosen sufficiently large and n is fixed we have 2B 1 > .kC1/Š nkŠ 10 10 contradicting the equality from the mean value theorem. Therefore c is transcendental. In 1873 Hermite proved that e is transcendental while Lindemann in 1882 showed that  is transcendental. Schneider in 1934 showed that ab is transcendental if a ¤ 0, a and b are algebraic and b is irrational. Later in the book we will prove that both e and  are transcendental. An interesting open question is the following: Is  transcendental over Q.e/? To close this section we show that in general if a 2 L is transcendental over K then K.a/jK is isomorphic to the field of rational functions over K.

78

Chapter 5 Field Extensions

Theorem 5.5.5. Suppose that LjK is a field extension and a 2 L is transcendental over K. Then K.a/jK is isomorphic to K.x/jK. Here the isomorphism  W K.x/ ! K.a/ can be chosen such that .x/ D a. Proof. Define the map  W K.x/ ! K.a/ by   f .a/ f .x/ D  g.x/ g.a/ for f .x/; g.x/ 2 KŒx with g.x/ ¤ 0. Then  is a homomorphism and .x/ D a. Since  ¤ 0 it follows that  is an isomorphism.

5.6 1.

Exercises Let a 2 C with a3  2a C 2 D 0 and b D a2  a. Compute the minimal polynomial mb .x/ of b over Q and compute the inverse of b in Q.a/.

Determine the algebraic closure of R in C.x/. p 2n 2 2 R, n D 1; 2; 3; : : : and A WD ¹an W n 2 Nº and E WD Q.A/. 3. Let an WD Show:

2.

(i) jQ.an / W Qj D 2n . (ii) jE W Qj D 1. S (iii) E D 1 nD1 Q.an /. (iv) E is algebraic over Q. 4.

Determine jE W Qj for p p (i) E D Q. 2; 2/. p p p (ii) E D Q. 3; 3 C 3 3/.

p ; 1Ci p /. (iii) E D Q. 1Ci 2 2 p p p p p 5. Show that Q. 2;p 3/pD ¹a C b 2 C c 3 C d 6 W a; b;p c; d p 2 Qº. Determine p 2; 3/ over Q. Further show that Q. 2; 3/ D Q. 2 C the degree of Q. p 3/.

6.

Let K, E be fields and a 2 E be transcendental over K. Show: (i) Each element of K.a/jK is transcendental over K. (ii) an is transcendental over K for each n > 1. 3

a (iii) If L WD K. aC1 / then a is algebraic over L. Determine the minimal polynomial ma .x/ of a over L.

79

Section 5.6 Exercises

7.

Let K be a field and a 2 K.x/jK. Show: (i) x is algebraic over K.a/. (ii) If L is a field with K  L K.x/, then jK.x/ W Lj < 1. (iii) a is transcendental over K.

8.

Suppose that a 2 L is algebraic over K. Let I D ¹f .x/ 2 KŒx W f .a/ D 0º: Since a is algebraic I ¤ ;. Prove that I is an ideal in KŒx.

9.

Prove that there are uncountably many transcendental numbers. To do this show that the set A of algebraic numbers is countable. To do this: (i) Show that Qn Œx the set of rational polynomials of degree  n is countable (finite Cartesian product of countable sets). (ii) Let Bn D ¹Zeros of polynomials in Qn º. Show that B is countable. S (iii) Show that A D 1 nD1 Bn and conclude that A is countable. (iv) Show that the transcendental numbers are uncountable.

10. Consider the map  W KŒx ! KŒa given by X  X i  ki x D ki a i : i

i

Show that  is a ring epimorphism. 11. Suppose that K is a field and R is an integral domain with K  R. Then R can be viewed as a vector space over K. Let r0 2 R with r0 ¤ 0. Define the map from R to R given by  .r/ D rr0 : Show that this is a linear transformation from R to R considered as a vector space over K.

Chapter 6

Field Extensions and Compass and Straightedge Constructions

6.1

Geometric Constructions

Greek mathematicians in the classical period posed the problem of constructing certain geometric figures in the Euclidean plane using only a straightedge and a compass. These are known as geometric construction problems. Recall from elementary geometry that using a straightedge and compass it is possible to draw a line parallel to a given line segment through a given point, to extend a given line segment, and to erect a perpendicular to a given line at a given point on that line. There were other geometric construction problems that the Greeks could not determine straightedge and compass solutions but on the other hand were never able to prove that such constructions were impossible. In particular there were four famous insolvable (to the Greeks) construction problems. The first is the squaring of the circle. This problem is, given a circle, to construct using straightedge and compass a square having area equal to that of the given circle. The second is the doubling of the cube. This problem is given a cube of given side length, to construct using a straightedge and compass, a side of a cube having double the volume of the original cube. The third problem is the trisection of an angle. This problem is to trisect a given angle using only a straightedge and compass. The final problem is the construction of a regular n-gon. This problems asks which regular n-gons could be constructed using only straightedge and compass. By translating each of these problems into the language of field extensions we can show that each of the first three problems are insolvable in general and we can give the complete solution to the construction of the regular n-gons.

6.2

Constructible Numbers and Field Extensions

We now translate the geometric construction problems into the language of field extensions. As a first step we define a constructible number. Definition 6.2.1. Suppose we are given a line segment of unit length. An ˛ 2 R is constructible if we can construct a line segment of length j˛j in a finite number of steps from the unit segment using a straightedge and compass.

Section 6.2 Constructible Numbers and Field Extensions

81

Our first result is that the set of all constructible numbers forms a subfield of R. Theorem 6.2.2. The set C of all constructible numbers forms a subfield of R. Further, Q  C . Proof. Let C be the set of all constructible numbers. Since the given unit length segment is constructible, we have 1 2 C. Therefore, C ¤ ;, and thus to show that it is a field we must show that it is closed under the field operations. Suppose ˛; ˇ are constructible. We must show then that ˛ ˙ ˇ; ˛ˇ, and ˛=ˇ for ˇ ¤ 0 are constructible. If ˛; ˇ > 0, construct a line segment of length j˛j. At one end of this line segment extend it by a segment of length jˇj. This will construct a segment of length ˛ C ˇ. Similarly, if ˛ > ˇ, lay off a segment of length jˇj at the beginning of a segment of length j˛j. The remaining piece will be ˛  ˇ. By considering cases we can do this in the same manner if either ˛ or ˇ or both are negative. These constructions are pictured in Figure 6.1. Therefore, ˛ ˙ ˇ are constructible.

Figure 6.1 In Figure 6.2 we show how to construct ˛ˇ. Let the line segment OA have length j˛j. Consider a line L through O not coincident with OA. Let OB have length jˇj as in the diagram. Let P be on ray OB so that OP has length 1. Draw AP and then find Q on ray OA such that BQ is parallel to AP . From similar triangles we then have jOP j jOBj

D

jOAj jOQj

)

j˛j 1 D : jˇj jOQj

Then jOQj D j˛jjˇj, and so ˛ˇ is constructible.

Figure 6.2

82

Chapter 6 Field Extensions and Compass and Straightedge Constructions

A similar construction, pictured in Figure 6.3, shows that ˛=ˇ for ˇ ¤ 0 is constructible. Find OA; OB; OP as above. Now, connect A to B and let PQ be parallel to AB. From similar triangles again we have jOQj j˛j 1 D H) D jOQj: jˇj j˛j jˇj Hence ˛=ˇ is constructible.

Figure 6.3 Therefore, C is a subfield of R. Since char C D 0, it follows that Q  C . Let us now consider analytically how a constructible number is found in the plane. Starting at the origin and using the unit length and the constructions above, we can locate any point in the plane with rational coordinates. That is, we can construct the point P D .q1 ; q2 / with q1 ; q2 2 Q. Using only straightedge and compass, any further point in the plane can be determined in one of the following three ways. 1. The intersection point of two lines each of which passes through two known points each having rational coordinates. 2. The intersection point of a line passing through two known points having rational coordinates and a circle whose center has rational coordinates and whose radius squared is rational. 3. The intersection point of two circles each of whose centers has rational coordinates and each of whose radii is the square root of a rational number. Analytically, the first case involves the solution of a pair of linear equations each with rational coefficients and thus only leads to other rational numbers. In cases two and three we must solve equations of the form x 2 C y 2 C ax C by C c D 0, with a; b; c 2 Q. These will then be quadratic equations over Q, and thus the solutions p will either be in Q or in a quadratic extension Q. ˛/ of Q. Once a real quadratic extension of Q is found, the process can be iterated. Conversely it can be shown that p if ˛ is constructible, so is ˛. We thus can prove the following theorem.

Section 6.3 Four Classical Construction Problems

83

Theorem 6.2.3. If is constructible with … Q, then there exists a finite number of elements ˛1 ; : : : ; ˛r 2 R with ˛r D such that for i D 1; : : : ; r, Q.˛1 ; : : : ; ˛i / is a quadratic extension of Q.˛1 ; : : : ; ˛i 1 /. In particular, jQ. / W Qj D 2n for some n  1. Therefore, the constructible numbers are precisely those real numbers that are contained in repeated quadratic extensions of Q. In the next section we use this idea to show the impossibility of the first three mentioned construction problems.

6.3

Four Classical Construction Problems

We now consider the aforementioned construction problems. Our main technique will be to use Theorem 6.2.3. From this result we have that if is constructible with

… Q, then jQ. / W Qj D 2n for some n  1.

6.3.1 Squaring the Circle Theorem 6.3.1. It is impossible to square the circle. That is, it is impossible in general, given a circle, to construct using straightedge and compass a square having area equal to that of the given circle. Proof. Suppose the given circle has radius 1. It is then constructible and would have p an area of . A corresponding square would then have to have a side of length . To be constructible a number ˛ must have jQ.˛/ W Qj D 2m < 1 and hence ˛ must p be algebraic. However  is transcendental, so  is also transcendental and therefore not constructible.

6.3.2 The Doubling of the Cube Theorem 6.3.2. It is impossible to double the cube. This means that it is impossible in general, given a cube of given side length, to construct using a straightedge and compass, a side of a cube having double the volume of the original cube. Proof. Let the given side length be 1, so that the original volume is also 1. To double this we would have to construct a side of length 21=3 . However jQ.21=3 / W Qj D 3 since the minimal polynomial over Q is m21=3 .x/ D x 3  2. This is not a power of 2 so 21=3 is not constructible.

6.3.3 The Trisection of an Angle Theorem 6.3.3. It is impossible to trisect an angle. This means that it is impossible in general to trisect a given angle using only a straightedge and compass.

84

Chapter 6 Field Extensions and Compass and Straightedge Constructions

Proof. An angle  is constructible if and only if a segment of length jcos j is constructible. Since cos.=3/ D 1=2, therefore =3 is constructible. We show that it cannot be trisected by straightedge and compass. The following trigonometric identity holds cos.3 / D 4 cos3 . /  3 cos. /: Let ˛ D cos.=9/. From the above identity we have 4˛ 3  3˛  12 D 0. The polynomial 4x 3  3x  12 is irreducible over Q, and hence the minimal polynomial over Q is m˛ .x/ D x 3  34 x  18 . It follows that jQ.˛/ W Qj D 3, and hence ˛ is not constructible. Therefore, the corresponding angle =9 is not constructible. Therefore, =3 is constructible, but it cannot be trisected.

6.3.4 Construction of a Regular n-Gon The final construction problem we consider is the construction of regular n-gons. The algebraic study of the constructibility of regular n-gons was initiated by Gauss in the early part of the nineteenth century. Notice first that a regular n-gon will be constructible for n  3 if and only if 2 the angle 2 n is constructible, which is the case if and only if the length cos n is a 2 constructible number. From our techniques if cos n is a constructible number then m necessarily jQ.cos. 2 n // W Qj D 2 for some m. After we discuss Galois theory we see that this condition is also sufficient. Therefore cos 2 n is a constructible number if 2 m and only if jQ.cos. n // W Qj D 2 for some m. The solution of this problem, that is the determination of when jQ.cos. 2 n / W Qj D m 2 involves two concepts from number theory; the Euler phi-function and Fermat primes. Definition 6.3.4. For any natural number n, the Euler phi-function is defined by .n/ D number of integers less than or equal to n and relatively prime to n: Example 6.3.5. .6/ D 2 since among 1; 2; 3; 4; 5; 6 only 1; 5 are relatively prime to 6. It is fairly straightforward to develop a formula for .n/. A formula is first determined for primes and for prime powers and then pasted back together via the fundamental theorem of arithmetic. Lemma 6.3.6. For any prime p and m > 0,   1 : .p m / D p m  p m1 D p m 1  p

85

Section 6.3 Four Classical Construction Problems

Proof. If 1  a  p then either a D p or .a; p/ D 1. It follows that the positive integers less than or equal to p m which are not relatively prime to p m are precisely the multiples of p that is p; 2p; 3p; : : : ; p m1  p. All other positive a < p m are relatively prime to p m . Hence the number relatively prime to p m is p m  p m1 : Lemma 6.3.7. If .a; b/ D 1 then .ab/ D .a/.b/. Proof. Given a natural number n a reduced residue system modulo n is a set of integers x1 ; : : : ; xk such that each xi is relatively prime to n, xi ¤ xj mod n unless i D j and if .x; n/ D 1 for some integer x then x  xi mod n for some i. Clearly .n/ is the size of a reduced residue system modulo n. Let Ra D ¹x1 ; : : : ; x.a/ º be a reduced residue system modulo a, Rb D ¹y1 ; : : : ; y.b/ º be a reduced residue system modulo b, and let S D ¹ayi C bxj W i D 1; : : : ; .b/; j D 1; : : : ; .a/º: We claim that S is a reduced residue system modulo ab. Since S has .a/.b/ elements it will follow that .ab/ D .a/.b/. To show that S is a reduced residue system modulo ab we must show three things: first that each x 2 S is relatively prime to ab; second that the elements of S are distinct; and finally that given any integer n with .n; ab/ D 1 then n  s mod ab for some s 2 S . Let x D ayi C bxj . Then since .xj ; a/ D 1 and .a; b/ D 1 it follows that .x; a/ D 1. Analogously .x; b/ D 1. Since x is relatively prime to both a and b we have .x; ab/ D 1. This shows that each element of S is relatively prime to ab. Next suppose that ayi C bxj  ayk C bxl

mod ab:

Then abj.ayi C bxj /  .ayk C bxl / H) ayi  ayk

mod b:

Since .a; b/ D 1 it follows that yi  yk mod b. But then yi D yk since Rb is a reduced residue system. Similarly xj D xl . This shows that the elements of S are distinct modulo ab. Finally suppose .n; ab/ D 1. Since .a; b/ D 1 there exist x; y with ax C by D 1. Then anx C bny D n: Since .x; b/ D 1 and .n; b/ D 1 it follows that .nx; b/ D 1. Therefore there is an si with nx D si C t b. In the same manner .ny; a/ D 1 and so there is an rj with

86

Chapter 6 Field Extensions and Compass and Straightedge Constructions

ny D rj C ua. Then a.si C t b/ C b.rj C ua/ D n H) n D asi C brj C .t C u/ab H) n  ari C bsj

mod ab

and we are done. We now give the general formula for .n/. e Theorem 6.3.8. Suppose n D p1e1    pkk then

.n/ D .p1e1  p1e1 1 /.p2e2  p2e2 1 /    .pkek  pkek 1 /: Proof. From the previous lemma we have e

.n/ D .p1e1 /.p2e2 /    .pkk / D .p1e1  p1e1 1 /.p2e2  p2e2 1 /    .pkek  pkek 1 / D p1e1 .1  1=p1 /    pkek .1  1=pk / D p1e1    pkek  .1  1=p1 /    .1  1=pk / Y D n .1  1=pi /: i

Example 6.3.9. Determine .126/. Now 126 D 2  32  7 H) .126/ D .2/.32 /.7/ D .1/.32  3/.6/ D 36: Hence there are 36 units in Z126 . An interesting result with many generalizations in number theory is the following. Theorem 6.3.10. For n > 1 and for d  1 X .d / D n: d jn

Proof. We first prove the theorem for prime powers and then paste together via the fundamental theorem of arithmetic. Suppose that n D p e for p a prime. Then the divisors of n are 1; p; p 2 ; : : : ; p e , so X .d / D .1/ C .p/ C .p 2 / C    C .p e / d jn

D 1 C .p  1/ C .p 2  p/ C    C .p e  p e1 /: Notice that this sum telescopes, that is 1 C .p  1/ D p; p C .p 2  p/ D p 2 and so on. Hence the sum is just p e and the result is proved for n a prime power.

87

Section 6.3 Four Classical Construction Problems

We now do an induction on the number of distinct prime factors of n. The above argument shows that the result is true if n has only one distinct prime factor. Assume that the result is true whenever an integer has less than k distinct prime factors and suppose n D p1e1    pkek has k distinct prime factors. Then n D p e c where p D p1 , e D e1 and c has fewer than k distinct prime factors. By the inductive hypothesis X

.d / D c:

d jc

Since .c; p/ D 1 the divisors of n are all of the form p ˛ d1 where d1 jc and ˛ D 0; 1; : : : ; e. It follows that X X X X .d / D .d1 / C .pd1 / C    C .p e d1 /: d jn

d1 jc

d1 jc

d1 jc

Since .d1 ; p ˛ / D 1 for any divisor of c this sum equals X

.d1 / C

d1 jc

D

X

X

.p/.d1 / C    C

d1 jc

.d1 / C .p  1/

d1 jc

X

X

.p e /.d1 /

d1 jc

.d1 / C    C .p e  p e1 /

d1 jc 2

X

.d1 /

d1 jc e

D c C .p  1/c C .p  p/c C    C .p  p

e1

/c:

As in the case of prime powers this sum telescopes giving a final result X

.d / D p e c D n:

d jn

Example 6.3.11. Consider n D 10. The divisors are 1; 2; 5; 10. Then .1/ D 1, .2/ D 1, .5/ D 4, .10/ D 4. Then .1/ C .2/ C .5/ C .10/ D 1 C 1 C 4 C 4 D 10: We will see later in the book that the Euler phi-function plays an important role in the structure theory of abelian groups. We now turn to Fermat primes. Definition 6.3.12. The Fermat numbers are the sequence .Fn / of positive integers defined by n Fn D 22 C 1; n D 0; 1; 2; 3; : : : : If a particular Fn is prime it is called a Fermat prime.

88

Chapter 6 Field Extensions and Compass and Straightedge Constructions

Fermat believed that all the numbers in this sequence were primes. In fact F0 ; F1 ; F2 ; F3 ; F4 are all prime but F5 is composite and divisible by 641 (see exercises). It is still an open question whether or not there are infinitely many Fermat primes. It has been conjectured that there are only finitely many. On the other hand if a number of the form 2n C 1 is a prime for some integer n then it must be a Fermat prime. Theorem 6.3.13. If a  2 and an C 1, n  1, is a prime then a is even and n D 2m for some nonnegative integer m. In particular if p D 2k C 1, k  1, is a prime then k D 2n for some n and p is a Fermat prime. Proof. If a is odd then an C 1 is even and hence not a prime. Suppose then that a is even and n D kl with k odd and k  3. Then akl C 1 D a.k1/l  a.k2/l C    C 1: al C 1 Therefore al C 1 divides akl C 1 if k  3. Hence if an C 1 is a prime we must have n D 2m . We can now state the solution to the constructibility of regular n-gons. Theorem 6.3.14. A regular n-gon is constructible with a straightedge and compass if and only if n D 2m p1    pk where p1 ; : : : ; pk are distinct Fermat primes. Before proving the theorem notice for example that a regular 20-gon is constructible since 20 D 22  5 and 5 is a Fermat prime. On the other hand a regular 11-gon is not constructible. Proof. Let  D e

2 i n

be a primitive n-th root of unity. Since     2 i 2 2 e n D cos C i sin n n

is easy to compute that (see exercises)   2 1 :  C D 2 cos  n Therefore Q. C we will prove that

1 /

D Q.cos. 2 n //. After we discuss Galois theory in more detail ˇ ˇ   ˇ ˇ ˇQ  C 1 W Qˇ D .n/ ˇ ˇ  2

where .n/ is the Euler phi-function. Therefore cos. 2 n / is constructible if and only .n/ if 2 and hence .n/ is a power of 2.

89

Section 6.4 Exercises

Suppose that n D 2m p1e1    pkek , all pi odd primes. Then from Theorem 6.3.8 .n/ D 2m1  .p1e1  p1e1 1 /.p2e2  p2e2 1 /    .pkek  pkek 1 /: If this was a power of 2 each factor must also be a power of 2. Now piei  piei 1 D piei 1 .pi  1/: If this is to be a power of 2 we must have ei D 1 and pi  1 D 2ki for some ki . Therefore each prime is distinct to the first power and pi D 2ki C 1 is a Fermat prime proving the theorem.

6.4

Exercises

1. Let  be a given angle. In which of the following cases is the angle from the angle  by compass and straightedge? (a)  D (b)  D (c)  D

 13 ,  33 ,  7,

constructible

 26 .  D 11 .  D 12 .

D

2. (The golden section) In the plane let AB be a given segment from A to B with length a. The segment AB should be divided such that the proportion of AB to the length of the bigger subsegment is equal to the proportion of the length of the bigger subsegment to the length of the smaller subsegment: b a D ; b ab where b is the length of the bigger subsegment. Such a division is called division x by the golden section. If we write b D ax, 0 < x < 1, then x1 D 1x , that is 2 D 1  x. Show: x (a)

1 x

D

p 1C 5 2

D ˛.

(b) Construct the division of AB by the golden section with compass and straightedge. (c) If we divide the radius r > 0 of a circle by the golden section, then the bigger part of the so divided radius is the side of the regular 10-gon with its 10 vertices on the circle. 3. Given a regular 10-gon such that the 10 vertices are on the circle with radius R > 0. Show that the length of each side is equal to the bigger part of the, by the golden section divided, radius. Describe the procedure of the construction of the regular 10-gon and 5-gon.

90

Chapter 6 Field Extensions and Compass and Straightedge Constructions

4. Construct the regular 17-gon with compass and straightedge. Hint: We have to 2 i 17 . First, construct construct the number 12 .! C ! 1 / D cos 2 17 , where ! D e 2 the positive zero !1 of the polynomial x C x  4; we get 1 p !1 D . 17  1/ D ! C ! 1 C ! 2 C ! 2 C ! 4 C ! 4 C ! 8 C ! 8 : 2 Then, construct the positive zero !2 of the polynomial x 2  !1 x  1; we get q p  1 p 17  1 C 34  2 17 D ! C ! 1 C ! 4 C ! 4 : !2 D 4 From !1 and !2 construct ˇ D 12 .!22  !1 C !2  4/. Then !3 D 2 cos 2 17 is the biggest of the two positive zeros of the polynomial x 2  !2 x C ˇ. 5. The Fibonacci-numbers fn , n 2 N [ ¹0º, are defined by f0 D 0, f1 D 1 and fnC2 D fnC1 C fn for n 2 N [ ¹0º. Show: (a) fn D

˛ n ˇ n ˛ˇ

with ˛ D

p 1C 5 2 ,

ˇD

f

(b) . nC1 / converges and limn!1 fn n2N  0 1 n  fn1 fn  (c) 1 1 D fn fnC1 , n 2 N.

p 1 5 2 .

fnC1 fn

D

p 1C 5 2

D ˛.

(d) f1 C f2 C    C fn D fnC2  1, n  1. (e) fn1 fnC1  fn2 D .1/n , n 2 N. (f) f12 C f22 C    C fn2 D fn fnC1 , n 2 N. (g) The Fermat numbers F0 ; F1 ; F2 ; F3 ; F4 . are all prime but F5 is composite and divisible by 641. 6. Let  D e

show that

2 i n

be a primitive n-th root of unity. Using     2 i 2 2 e n D cos C i sin n n   2 1  C D 2 cos :  n

Chapter 7

Kronecker’s Theorem and Algebraic Closures

7.1

Kronecker’s Theorem

In the last chapter we proved that if LjK is a field extension then there exists an intermediate field K  A  L such that A is algebraic over K and contains all the elements of L that are algebraic over K. We call A the algebraic closure of K within L. In this chapter we prove that starting with any field K we can construct an extension field K that is algebraic over K and is algebraically closed. By this we mean that there are no algebraic extensions of K or equivalently that there are no irreducible nonlinear polynomials in KŒx. In the final section of this chapter we will give a proof of the famous fundamental theorem of algebra which in the language of this chapter says that the field C of complex numbers is algebraically closed. We will present another proof of this important result later in the book after we discuss Galois theory. First we need the following crucial result of Kronecker which says that given a polynomial f .x/ in KŒx where K is a field we can construct an extension field L of K in which f .x/ has a root ˛. We say that L has been constructed by adjoining ˛ to K. Recall that if f .x/ 2 KŒx is irreducible then f .x/ can have no roots in K. We first need the following concept. Definition 7.1.1. Let LjK and L0 jK be field extensions. Then a K-isomorphism is an isomorphism  W L ! L0 that is the identity map on K, that is fixes each element of K. Theorem 7.1.2 (Kronecker’s theorem). Let K be a field and f .x/ 2 KŒx. Then there exists a finite extension K 0 of K where f .x/ has a root. Proof. Suppose that f .x/ 2 KŒx. We know that f .x/ factors into irreducible polynomials. Let p.x/ be an irreducible factor of f .x/. From the material in Chapter 4 we know that since p.x/ is irreducible the principal ideal hp.x/i in KŒx is a maximal ideal. To see this suppose that g.x/ … hp.x/i, so that g.x/ is not a multiple of p.x/. Since p.x/ is irreducible, it follows that .p.x/; g.x// D 1. Thus there exist h.x/; k.x/ 2 KŒx with h.x/p.x/ C k.x/g.x/ D 1:

92

Chapter 7 Kronecker’s Theorem and Algebraic Closures

The element on the left is in the ideal .g.x/; .p.x//, so the identity, 1, is in this ideal. Therefore, the whole ring KŒx is in this ideal. Since g.x/ was arbitrary, this implies that the principal ideal hp.x/i is maximal. Now let K 0 D KŒx=hp.x/i. Since hp.x/i is a maximal ideal it follows that K 0 is a field. We show that K can be embedded in K 0 and that p.x/ has a zero in K 0 . First consider the map ˛ W KŒx ! K 0 by ˛.f .x// D f .x/ C hp.x/i. This is a homomorphism. Since the identity element 1 2 K is not in hp.x/i it follows that ˛ restricted to K is nontrivial. Therefore ˛ restricted to K is a monomorphism since if ker.˛jK / ¤ K then ker.˛jK / D ¹0º. Therefore K can be embedded into ˛.K/ which is contained in K 0 . Therefore K 0 can be considered as an extension field of K. Consider the element a D x C hp.x/i 2 K 0 . Then p.a/ D p.x/ C hp.x/i D 0 C hp.x/i since p.x/ 2 hp.x/i. But 0 C hp.x/i is the zero element 0 of the factor ring KŒx=hp.x/i. Therefore in K 0 we have p.a/ D 0 and hence p.x/ has a zero in K 0 . Since p.x/ divides f .x/ we must have f .a/ D 0 in K 0 also. Therefore we have constructed an extension field of K in which f .x/ has a zero. We now outline a slightly more constructive proof of Kronecker’s theorem. From this construction we say that the field K 0 constructed by adjoining the root ˛ to K. Proof of Kronecker’s theorem. We can assume that f .x/ is irreducible. Suppose that f .x/ D a0 C a1 x C    C an x n with an ¤ 0. Define ˛ to satisfy a0 C a1 ˛ C    C an ˛ n D 0: Now define K 0 D K.˛/ in the following manner. We let K.˛/ D ¹c0 C c1 ˛ C    C cn1 ˛ n1 W ci 2 Kº: Then on K.˛/ define addition and subtraction componentwise and define multiplication by algebraic manipulation, replacing powers of ˛ higher than ˛ n by using ˛n D

a0  a1 ˛      an1 ˛ n1 : an

We claim that K 0 D K.˛/ then forms a field of finite degree over K. The basic ring properties follow easily by computation (see exercises) using the definitions. We must show then that every nonzero element of K.˛/ has a multiplicative inverse. Let g.˛/ 2 K.˛/. Then the corresponding polynomial g.x/ 2 KŒx is a polynomial of degree  n  1. Since f .x/ is irreducible of degree n it follows that f .x/ and g.x/ must be relatively prime, that is .f .x/; g.x// D 1. Hence there exist a.x/; b.x/ 2 KŒx with a.x/f .x/ C b.x/g.x/ D 1: Evaluate these polynomials at ˛ to get a.˛/f .˛/ C b.˛/g.˛/ D 1:

93

Section 7.1 Kronecker’s Theorem

Since by definition we have f .˛/ D 0 this becomes b.˛/g.˛/ D 1: Now b.˛/ might have degree higher than n  1 in ˛. However using the relation that f .˛/ D 0 we can rewrite b.˛/ as b.˛/ where b.˛/ now has degree  n  1 in ˛ and hence is in K.˛/. Therefore b.˛/g.˛/ D 1 and hence g.˛/ has a multiplicative inverse. It follows that K.˛/ is a field and by definition f .˛/ D 0. The elements 1; ˛; : : : ; ˛ n1 form a basis for K.˛/ over K and hence jK.˛/ W Kj D n: Example 7.1.3. Let f .x/ D x 2 C 1 2 RŒx. This is irreducible over R. We construct the field in which this has a root. Let ˛ be an indeterminate with ˛ 2 C 1 D 0 or ˛ 2 D 1. The extension field R.˛/ then has the form R.˛/ D ¹x C ˛y W x; y 2 R; ˛ 2 D 1º: It is clear that this field is R-isomorphic to the complex numbers C, that is, R.˛/ Š R.i/ Š C. In Chapter 5 we showed that if LjK is a field extension and a 2 L is algebraic over K then there is a smallest algebraic extension K.a/ of K within L. Further K.a/ is determined by the minimal polynomial ma .x/. The difference between this construction and the construction in Kronecker’s theorem is that in the proof of Kronecker’s theorem ˛ is defined to be the root and we constructed the field around it, whereas in the previous construction ˛ was assumed to satisfy the polynomial and K.˛/ was an already existing field that contained ˛. However the next theorem says that these constructions are the same up to K-isomorphism. Theorem 7.1.4. Let p.x/ 2 KŒx be an irreducible polynomial and let K 0 D K.˛/ be the extension field of K constructed in Kronecker’s theorem in which p.x/ has a zero ˛. Let L be an extension field of K and suppose that a 2 L is algebraic with minimal polynomial m˛ .x/ D p.x/. Then K.˛/ is K-isomorphic to K.a/. Proof. If LjK is a field extension and a 2 L is algebraic then the construction of the subfield K.a/ within L is identical to the construction outlined above for K.˛/. If deg.p.x// D n then the elements 1; a; : : : ; a n1 constitute a basis for K.a/ over K and the elements 1; ˛; : : : ; ˛ n1 constitute a basis for K.˛/ over K. The mapping  W K.a/ ! K.˛/ defined by  .k/ D k if k 2 K and  .a/ D ˛ and then extended by linearity is easily shown to be a K-isomorphism (see exercises).

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Theorem 7.1.5. Let K be a field. Then the following are equivalent. (1) Each nonconstant polynomial in KŒx has a zero in K. (2) Each nonconstant polynomial in KŒx factors into linear factors over K. That is, for each f .x/ 2 KŒx there exist elements a1 ; : : : ; an ; b 2 K with f .x/ D b.x  a1 /    .x  an /: (3) An element of KŒx is irreducible if and only if it is of degree one. (4) If LjK is an algebraic extension then L D K. Proof. Suppose that each nonconstant polynomial in KŒx has a zero in K. Let f .x/ 2 KŒx with deg.f .x// D n. Suppose that a1 is a zero of f .x/ then f .x/ D .x  a1 /h.x/ where the degree of h.x/ is n  1. Now h.x/ has a zero a2 in K so that f .x/ D .x  a1 /.x  a2 /g.x/ with deg.g.x// D n  2. Continue in this manner and f .x/ factors completely into linear factors. Hence (1) implies (2). Now suppose (2), that is that each nonconstant polynomial in KŒx factors into linear factors over K. Suppose that f .x/ is irreducible. If deg.f .x// > 1 then f .x/ factors into linear factors and hence is not irreducible. Therefore f .x/ must be of degree 1 and (2) implies (3). Now suppose that an element of KŒx is irreducible if and only if it is of degree one and suppose that LjK is an algebraic extension. Let a 2 L. Then a is algebraic over K. Its minimal polynomial ma .x/ is monic and irreducible over K and hence from (3) is linear. Therefore ma .x/ D x  a 2 KŒx. It follows that a 2 K and hence K D L. Therefore (3) implies (4). Finally suppose that whenever LjK is an algebraic extension then L D K. Suppose that f .x/ is a nonconstant polynomial in KŒx. From Kronecker’s theorem there exists a field extension L and a 2 L with f .a/ D 0. However L is an algebraic extension so by supposition K D L. Therefore a 2 K and f .x/ has a zero in K. Therefore (4) implies (1) completing the proof. In the next section we will prove that given a field K we can always find an extension field K with the properties of the last theorem.

7.2

Algebraic Closures and Algebraically Closed Fields

A field K is termed algebraically closed if K has no algebraic extensions other than K itself. This is equivalent to any one of the conditions of Theorem 7.1.5.

Section 7.2 Algebraic Closures and Algebraically Closed Fields

95

Definition 7.2.1. A field K is algebraically closed if every nonconstant polynomial f .x/ 2 KŒx has a zero in K. The following theorem is just a restatement of Theorem 7.1.5. Theorem 7.2.2. A field K is algebraically closed if and only it satisfies any one of the following conditions. (1) Each nonconstant polynomial in KŒx has a zero in K. (2) Each nonconstant polynomial in KŒx factors into linear factors over K. That is, for each f .x/ 2 KŒx there exist elements a1 ; : : : ; an ; b 2 K with f .x/ D b.x  a1 /    .x  an /: (3) An element of KŒx is irreducible if and only if it is of degree one. (4) If LjK is an algebraic extension then L D K. The prime example of an algebraically closed field is the field C of complex numbers. The fundamental theorem of algebra says that any nonconstant complex polynomial has a complex root. We now show that the algebraic closure of one field within an algebraically closed field is algebraically closed. First we define a general algebraic closure. Definition 7.2.3. An extension field K of a field K is an algebraic closure of K if K is algebraically closed and KjK is algebraic. Theorem 7.2.4. Let K be a field and LjK an extension of K with L algebraically closed. Let K D AK be the algebraic closure of K within L. Then K is an algebraic closure of K. Proof. Let K D AK be the algebraic closure of K within L. We know that KjK is algebraic therefore we must show that K is algebraically closed. Let f .x/ be a nonconstant polynomial in KŒx. Then f .x/ 2 LŒx. Since L is algebraically closed f .x/ has a zero a in L. Since f .a/ D 0 and f .x/ 2 KŒx it follows that a is algebraic over K. However K is algebraic over K and therefore a is also algebraic over K. Hence a 2 K and f .x/ has a zero in K. Therefore K is algebraically closed. We want to note the distinction between being algebraically closed and being an algebraic closure. Lemma 7.2.5. The complex numbers C is an algebraic closure of R but not an algebraic closure of Q. An algebraic closure of Q is A the field of algebraic numbers within C.

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Proof. C is algebraically closed (the fundamental theorem of algebra) and since jC W Rj D 2 it is algebraic over R. Therefore C is an algebraic closure of R. Although C is algebraically closed and contains the rational numbers Q it is not an algebraic closure of Q since it is not algebraic over Q since there exist transcendental elements. On the other hand, A, the field of algebraic numbers within Q, is an algebraic closure of Q from Theorem 7.2.4. We now show that every field has an algebraic closure. To do this we first show that any field can be embedded into an algebraically closed field. Theorem 7.2.6. Let K be a field. Then K can be embedded into an algebraically closed field. Proof. We show first that there is an extension field L of K in which each nonconstant polynomial f .x/ 2 KŒx has a zero in L. Assign to each nonconstant f .x/ 2 KŒx the symbol yf and consider R D KŒyf W f .x/ 2 KŒx the polynomial ring over K in the variables yf . Set I D

²X n

³ fj .yfj /rj W rj 2 R; fj .x/ 2 KŒx :

j D1

It is straightforward that I is an ideal in R. Suppose that I D R. Then 1 2 I . Hence there is a linear combination 1 D g1 f1 .yf1 / C    C gn fn .yfn / where gi 2 I D R. In the n polynomials g1 ; : : : ; gn there are only a finite number of variables, say for example yf1 ; : : : ; yfn ; : : : ; yfm : Hence 1D

n X

gi .yf1 ; : : : ; yfm /fi .yfi /. /:

i D1

Successive applications of Kronecker’s theorem lead us to construct an extension field P of K in which each fi has a zero ai . Substituting ai for yfi in (8) above we get that 1 D 0 a contradiction. Therefore I ¤ R. Since I is a ideal not equal to the whole ring R it follows that I is contained in a maximal ideal M of R. Set L D R=M . Since M is maximal K is a field. Now K \ M D ¹0º. If not suppose that a 2 K \ M with a ¤ 0. Then a1 a D 1 2 M

Section 7.2 Algebraic Closures and Algebraically Closed Fields

97

and then M D R. Now define  W K ! L by  .k/ D k C M . Since K \ M D ¹0º it follows that ker. / D ¹0º so  is a monomorphism. This allows us to identify K and  .K/ and shows that K embeds into L. Now suppose that f .x/ is a nonconstant polynomial in KŒx. Then f .yf C M / D f .yf / C M: However by the construction f .yf / 2 M so that f .yf C M / D M C M D the zero element of L: Therefore yf C M is a zero of f .x/. Therefore we have constructed a field L in which every nonconstant polynomial in KŒx has a zero in L. We now iterate this procedure to form a chain of fields K  K1 .D L/  K2     such that each nonconstant polynomial of Ki Œx has a zero in Ki C1 . S Now let KO D I Ki . It is easy to show (see exercises) that KO is a field. If f .x/ is O a nonconstant polynomial in KŒx then there is some i with f .x/ 2 Ki Œx. Therefore O Hence f .x/ has a zero in KO and KO is algebraically f .x/ has a zero in Ki C1 Œx  K. closed. Theorem 7.2.7. Let K be a field. Then K has an algebraic closure. Proof. Let KO be an algebraically closed field containing K which exists from Theorem 7.2.6. Now let K D AKO be the set of elements of KO that are algebraic over K. From Theorem 7.2.4 KO is an algebraic closure of K. The following lemma is straightforward. We leave the proof to the exercises. Lemma 7.2.8. Let K; K 0 be fields and  W K ! K 0 a homomorphism. Then Q W KŒx ! K 0 Œx given by  X X n n i Q ki x D ..ki //x i  i D1

i D0

Q If  is is also a homomorphism. By convention we identify  and Q and write  D . Q an isomorphism then so is .

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Lemma 7.2.9. Let K; K 0 be fields and  W K ! K 0 an isomorphism. Let f .x/ 2 KŒx be irreducible. Let K  K.a/ and K 0  K 0 .a0 / where a is a zero of f .x/ and a0 is a zero of .f .x//. Then there is an isomorphism W K.a/ ! K 0 .a0 / with 0 is uniquely determined. jK D  and .a/ D a . Further Proof. This is a generalized version of Theorem 7.1.4. If b 2 K.a/ then from the construction of K.a/ there is a polynomial g.x/ 2 KŒx with b D g.a/. Define a map W K.a/ ! K 0 .a0 / by .b/ D .g.x//.a0 /: We show that is an isomorphism. First is well-defined. Suppose that b D g.a/ D h.a/ with h.x/ 2 KŒx. Then .g  h/.a/ D 0. Since f .x/ is irreducible this implies that f .x/ D cma .x/ and since a is a zero of .g  h/.x/ then f .x/j.g  h/.x/. Then .f .x//j..g.x//  .h.x///: Since .f .x//.a0 / D 0 this implies that .g.x//.a0 / D .h.x//.a0 / and hence the map is well-defined. It is easy to show that is a homomorphism. Let b1 D g1 .a/, b2 D g2 .a/. Then b b1 1 D g1 g2 .a/. Hence .b1 b2 / D ..g1 g2 //.a0 / D .g1 /.a0 /.g2 /.a0 / D

.b1 / .b2 /:

In the same manner we have .b1 C b2 / D .b1 / C .b2 /. Now suppose that k 2 K so that k 2 KŒx is a constant polynomial. Then .k/ D ..k//.a0 / D .k/. Therefore restricted to K is precisely . As is not the zero mapping it follows that is a monomorphism. Finally since K.a/ is generated from K and a, and restricted to K is  it follows that is uniquely determined by  and .a/ D a0 . Hence is unique. Theorem 7.2.10. Let LjK be an algebraic extension. Suppose that L1 is an algebraically closed field and  is an isomorphism from K to K1  L1 . Then there exists a monomorphism from L to L1 with jK D .

Section 7.2 Algebraic Closures and Algebraically Closed Fields

99

Before we give the proof we note that the theorem gives the following diagram:

In particular the theorem can be applied to monomorphisms of a field K within an algebraic closure K of K. Specifically suppose that K  K where K is an algebraic closure of K and let ˛ W K ! K be a monomorphism with ˛.K/ D K. Then there exists an automorphism ˛  of K with ˛jK D ˛. Proof of Theorem 7.2.10. Consider the set M D ¹.M;  / W M is a field with K  M  L; where there exists a monomorphism  W M ! L1 with jK D º: Now the set M is nonempty since .K; / 2 M. Order M by .M1 ; 1 / < .M2 ; 2 / if M1  M2 and .2 /jM1 D 1 . Let K D ¹.Mi ; i / W i 2 I º be a chain in M. Let .M;  / be defined by M D

[

Mi

with  .a/ D i .a/ for all a 2 Mi :

i 2I

It is clear that M is an upper bound for the chain K. Since each chain has an upper bound it follows from Zorn’s lemma that M has a maximal element .N; /. We show that N D L. Suppose that N ¨ L. Let a 2 L n N . Then a is algebraic over N and further algebraic over K since LjK is algebraic. Let ma .x/ 2 N Œx be the minimal polynomial of a relative to N . Since L1 is algebraically closed .ma .x// has a zero a0 2 L1 . Therefore there is a monomorphism 0 W N.a/ ! L1 with 0 restricted to N the same as . It follows that .N; / < .N.a/; 0 / since a … N . This contradicts the maximality of N . Therefore N D L completing the proof. Combining the previous two theorems we can now prove that any two algebraic closures of a field K are unique up to K-isomorphism, that is up to an isomorphism that is the identity on K.

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Chapter 7 Kronecker’s Theorem and Algebraic Closures

Theorem 7.2.11. Let L1 and L2 be algebraic closures of the field K. Then there is a K-isomorphism  W L ! L1 . Again by K-isomorphism we mean that  is the identity on K. Proof. From Theorem 7.2.7 there is a monomorphism  W L1 ! L2 with  the identity on K. However since L1 is algebraically closed so is  .L1 /. Then L2 j .L1 / is an algebraic extension and since L2 is algebraically closed we must have L2 D  .L1 /. Therefore  is also surjective and hence an isomorphism. The following corollary is immediate. Corollary 7.2.12. Let LjK and L0 jK be field extensions with a 2 L and a0 2 L0 algebraic elements over K. Then K.a/ is K-isomorphic to K.a0 / if and only if jK.a/ W Kj D jK.a0 / W Kj and there is an element a00 2 K.a0 / with ma .x/ D ma00 .x/.

7.3

The Fundamental Theorem of Algebra

In this section we give a proof of the fact that the complex numbers form an algebraically closed field. This is known as the fundamental theorem of algebra. First we need the concept of a splitting field for a polynomial. In the next chapter we will examine this concept more deeply.

7.3.1 Splitting Fields We have just seen that given an irreducible polynomial over a field F we could always find a field extension in which this polynomial has a root. We now push this further to obtain field extensions where a given polynomial has all its roots. Definition 7.3.1. If K is a field and 0 ¤ f .x/ 2 KŒx and K 0 is an extension field of K, then f .x/ splits in K 0 (K 0 may be K), if f .x/ factors into linear factors in K 0 Œx. Equivalently, this means that all the roots of f .x/ are in K 0 . K 0 is a splitting field for f .x/ over K if K 0 is the smallest extension field of K in which f .x/ splits. (A splitting field for f .x/ is the smallest extension field in which f .x/ has all its possible roots.) K 0 is a splitting field over K if it is the splitting field for some finite set of polynomials over K. Theorem 7.3.2. If K is a field and 0 ¤ f .x/ 2 KŒx, then there exists a splitting field for f .x/ over K. Proof. The splitting field is constructed by repeated adjoining of roots. Suppose without loss of generality that f .x/ is irreducible of degree n over F . From Theorem 6.2.2 there exists a field F 0 containing ˛ with f .˛/ D 0. Then f .x/ D .x˛/g.x/ 2 F 0 Œx

Section 7.3 The Fundamental Theorem of Algebra

101

with deg g.x/ D n  1. By an inductive argument g.x/ has a splitting field and therefore so does f .x/. In the next chapter we will give a further characterization of splitting fields.

7.3.2 Permutations and Symmetric Polynomials To obtain a proof of the fundamental theorem of algebra we need to go a bit outside of our main discussions of rings and fields and introduce symmetric polynomials. In order to introduce this concept we first review some basic ideas from elementary group theory which we will look at in detail later in the book. Definition 7.3.3. A group G is a set with one binary operation which we will denote by multiplication, such that (1) The operation is associative, that is, .g1 g2 /g3 D g1 .g2 g3 / for all g1 ; g2 ; g3 2 G. (2) There exists an identity for this operation, that is, an element 1 such that 1g D g for each g 2 G. (3) Each g 2 G has an inverse for this operation, that is, for each g there exists a g 1 with the property that gg1 D 1. If in addition the operation is commutative (g1 g2 D g2 g1 for all g1 ; g2 2 G), the group G is called an abelian group. The order of G is the number of elements in G, denoted jGj. If jGj < 1; G is a finite group. H  G is a subgroup if H is also a group under the same operation as G. Equivalently, H is a subgroup if H ¤ ; and H is closed under the operation and inverses. Groups most often arise from invertible mappings of a set onto itself. Such mappings are called permutations. Definition 7.3.4. If T is a set, a permutation on T is a one-to-one mapping of T onto itself. We denote by ST the set of all permutations on T . Theorem 7.3.5. For any set T , ST forms a group under composition called the symmetric group on T . If T; T1 have the same cardinality (size), then ST Š ST1 . If T is a finite set with jT j D n, then ST is a finite group and jST j D nŠ. Proof. If ST is the set of all permutations on the set T , we must show that composition is an operation on ST that is associative and has an identity and inverses. Let f; g 2 ST . Then f; g are one-to-one mappings of T onto itself. Consider f ıg W T ! T . If f ıg.t1 / D f ıg.t2 /, then f .g.t1 // D f .g.t2 // and g.t1 / D g.t2 /, since f is one-to-one. But then t1 D t2 since g is one-to-one. If t 2 T , there exists t1 2 T with f .t1 / D t since f is onto. Then there exists t2 2 T with g.t2 / D t1 since g is onto. Putting these together, f .g.t2 // D t , and

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Chapter 7 Kronecker’s Theorem and Algebraic Closures

therefore f ı g is onto. Therefore, f ı g is also a permutation and composition gives a valid binary operation on ST . The identity function 1.t / D t for all t 2 T will serve as the identity for ST , while the inverse function for each permutation will be the inverse. Such unique inverse functions exist since each permutation is a bijection. Finally, composition of functions is always associative and therefore ST forms a group. If T; T1 have the same cardinality, then there exists a bijection W T ! T1 . Define a map F W ST ! ST1 in the following manner: if f 2 ST , let F .f / be the permutation on T1 given by F .f /.t1 / D .f . 1 .t1 ///. It is straightforward to verify that F is an isomorphism (see the exercises). Finally, suppose jT j D n < 1. Then T D ¹t1 ; : : : ; tn º. Each f 2 ST can be pictured as   t1 : : : t n : f D f .t1 / : : : f .tn / For t1 there are n choices for f .t1 /. For t2 there are only n  1 choices since f is one-to-one. This continues down to only one choice for tn . Using the multiplication principle, the number of choices for f and therefore the size of ST is n.n  1/    1 D nŠ: For a set with n elements we denote ST by Sn called the symmetric group on n symbols. Example 7.3.6. Write down the six elements of S3 and give the multiplication table for the group. Name the three elements 1; 2; 3. The six elements of S3 are then:       1 2 3 1 2 3 1 2 3 1D ; aD ; bD 1 2 3 2 3 1 3 1 2       1 2 3 1 2 3 1 2 3 cD ; dD ; eD : 2 1 3 3 2 1 1 3 2 The multiplication table for S3 can be written down directly by doing the required composition. For example,      1 2 3 1 2 3 1 2 3 ac D D D d: 2 3 1 2 1 3 3 2 1 To see this, note that a W 1 ! 2; 2 ! 3; 3 ! 1; c W 1 ! 2; 2 ! 1; 3 ! 3 and so ac W 1 ! 3; 2 ! 2; 3 ! 1. It is somewhat easier to construct the multiplication table if we make some observations. First, a2 D b and a3 D 1. Next, c 2 D 1, d D ac, e D a2 c and finally ac D ca2 .

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Section 7.3 The Fundamental Theorem of Algebra

From these relations the following multiplication table can be constructed:

1 a a2 c ac a2 c

1 1 a a2 c ac a2 c

a a a2 1 a2 c c ac

a2 a2 1 a ac a2 c c

c c ac a2 c 1 a a2

ac ac a2 c c a2 1 a

a2 c a2 c c ac : a a2 1

To see this, consider, for example, .ac/a2 D a.ca2 / D a.ac/ D a2 c. More generally, we can say that S3 has a presentation given by S3 D ha; cI a3 D c 2 D 1; ac D ca2 i: By this we mean that S3 is generated by a; c, or that S3 has generators a; c and the whole group and its multiplication table can be generated by using the relations a3 D c 2 D 1, ac D ca2 . An important result, the form of which we will see later in our work on extension fields, is the following. Lemma 7.3.7. Let T be a set and T1  T a subset. Let H be the subset of ST that fixes each element of T1 , that is, f 2 H if f .t / D t for all t 2 T1 . Then H is a subgroup. Proof. H ¤ ; since 1 2 H . Now suppose h1 ; h2 2 H . Let t1 2 T1 and consider h1 ı h2 .t1 / D h1 .h2 .t1 //. Now h2 .t1 / D t1 since h2 2 H , but then h1 .t1 / D t1 since h1 2 H . Therefore, h1 ı h2 2 H and H is closed under composition. If h1 fixes t1 then h1 1 also fixes t1 so H is also closed under inverses and is therefore a subgroup. We now apply these ideas of permutations to certain polynomials in independent indeterminates over a field. We will look at these in detail later in this book. Definition 7.3.8. Let y1 ; : : : ; yn be (independent) indeterminates over a field K. A polynomial f .y1 ; : : : ; yn / 2 KŒy1 ; : : : ; yn  is a symmetric polynomial in y1 ; : : : ; yn if f .y1 ; : : : ; yn / is unchanged by any permutation of ¹y1 ; : : : ; yn º, that is, f .y1 ; : : : ; yn / D f . .y1 /; : : : ; .yn //. If K  K 0 are fields and ˛1 ; : : : ; ˛n are in F 0 , then we call a polynomial f .˛1 ; : : : ; ˛n / with coefficients in K symmetric in ˛1 ; : : : ; ˛n if f .˛1 ; : : : ; ˛n / is unchanged by any permutation of ¹˛1 ; : : : ; ˛n º.

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Chapter 7 Kronecker’s Theorem and Algebraic Closures

Example 7.3.9. Let K be a field and k0 ; k1 2 K. Let h.y1 ; y2 / D k0 .y1 C y2 / C k1 .y1 y2 /. There are two permutations on ¹y1 ; y2 º, namely 1 W y1 ! y1 , y2 ! y2 and

2 W y1 ! y2 , y2 ! y1 . Applying either one of these two to ¹y1 ; y2 º leaves h.y1 ; y2 / invariant. Therefore, h.y1 ; y2 / is a symmetric polynomial. Definition 7.3.10. Let x; y1 ; : : : ; yn be indeterminates over a field K (or elements of an extension field K 0 of K). Form the polynomial p.x; y1 ; : : : ; yn / D .x  y1 /    .x  yn /: The i-th elementary symmetric polynomial si in y1 ; : : : ; yn for i D 1; : : : ; n, is .1/i ai , where ai is the coefficient of x ni in p.x; y1 ; : : : ; yn /. Example 7.3.11. Consider y1 ; y2 ; y3 . Then p.x; y1 ; y2 ; y3 / D .x  y1 /.x  y2 /.x  y3 / D x 3  .y1 C y2 C y3 /x 2 C .y1 y2 C y1 y3 C y2 y3 /x  y1 y2 y3 : Therefore, the three elementary symmetric polynomials in y1 ; y2 ; y3 over any field are (1) s1 D y1 C y2 C y3 . (2) s2 D y1 y2 C y1 y3 C y2 y3 . (3) s3 D y1 y2 y3 . In general, the pattern of the last example holds for y1 ; : : : ; yn . That is, s1 D y1 C y2 C    C yn s2 D y1 y2 C y1 y3 C    C yn1 yn s3 D y1 y2 y3 C y1 y2 y4 C    C yn2 yn1 yn :: : sn D y1    yn : The importance of the elementary symmetric polynomials is that any symmetric polynomial can be built up from the elementary symmetric polynomials. We make this precise in the next theorem called the fundamental theorem of symmetric polynomials. We will use this important result several times, and we will give a complete proof in Section 7.5.

Section 7.4 The Fundamental Theorem of Algebra

105

Theorem 7.3.12 (fundamental theorem of symmetric polynomials). If P is a symmetric polynomial in the indeterminates y1 ; : : : ; yn over a field K, that is, P 2 KŒy1 ; : : : ; yn  and P is symmetric, then there exists a unique g 2 KŒy1 ; : : : ; yn  such that f .y1 ; : : : ; yn / D g.s1 ; : : : ; sn /. That is, any symmetric polynomial in y1 ; : : : ; yn is a polynomial expression in the elementary symmetric polynomials in y1 ; : : : ; yn . From this theorem we obtain the following two lemmas, which will be crucial in our proof of the fundamental theorem of algebra. Lemma 7.3.13. Let p.x/ 2 KŒx and suppose p.x/ has the roots ˛1 ; : : : ; ˛n in the splitting field K 0 . Then the elementary symmetric polynomials in ˛1 ; : : : ; ˛n are in K. Proof. Suppose p.x/ D c0 C c1 x C    C cn x n 2 KŒx. In K 0 Œx, p.x/ splits, with roots ˛1 ; : : : ; ˛n , and thus in K 0 Œx, p.x/ D cn .x  ˛1 /    .x  ˛n /: The coefficients are then cn .1/i si .˛1 ; : : : ; ˛n /, where the si .˛1 ; : : : ; ˛n / are the elementary symmetric polynomials in ˛1 ; : : : ; ˛n . However, p.x/ 2 KŒx, so each coefficient is in K. It follows then that for each i, cn .1/i si .˛1 ; : : : ; ˛n / 2 K, and hence si .˛1 ; : : : ; ˛n / 2 K since cn 2 K. Lemma 7.3.14. Let p.x/ 2 KŒx and suppose p.x/ has the roots ˛1 ; : : : ; ˛n in the splitting field K 0 . Suppose further that g.x/ D g.x; ˛1 ; : : : ; ˛n / 2 K 0 Œx. If g.x/ is a symmetric polynomial in ˛1 ; : : : ; ˛n , then g.x/ 2 KŒx. Proof. If g.x/ D g.x; ˛1 ; : : : ; ˛n / is symmetric in ˛1 ; : : : ; ˛n , then from Theorem 7.3.12 it is a symmetric polynomial in the elementary symmetric polynomials in ˛1 ; : : : ; ˛n . From Lemma 7.3.13 these are in the ground field K, so the coefficients of g.x/ are in K. Therefore, g.x/ 2 KŒx.

7.4

The Fundamental Theorem of Algebra

We now present a proof of the fundamental theorem of algebra. Theorem 7.4.1 (fundamental theorem of algebra). Any nonconstant complex polynomial has a complex root. In other words, the complex number field C is algebraically closed. The proof depends on the following sequence lemmas. The crucial one now is the last, which says that any real polynomial must have a complex root. Lemma 7.4.2. Any odd-degree real polynomial must have a real root.

106

Chapter 7 Kronecker’s Theorem and Algebraic Closures

Proof. This is a consequence of the intermediate value theorem from analysis. Suppose P .x/ 2 RŒx with deg P .x/ D n D 2k C 1 and suppose the leading coefficient an > 0 (the proof is almost identical if an < 0). Then P .x/ D an x n C (lower terms) and n is odd. Then, (1) limx!1 P .x/ D limx!1 an x n D 1 since an > 0. (2) limx!1 P .x/ D limx!1 an x n D 1 since an > 0 and n is odd. From (1), P .x/ gets arbitrarily large positively, so there exists an x1 with P .x1 / > 0. Similarly, from (2) there exists an x2 with P .x2 / < 0. A real polynomial is a continuous real-valued function for all x 2 R. Since P .x1 /P .x2 / < 0, it follows from the intermediate value theorem that there exists an x3 , between x1 and x2 , such that P .x3 / D 0. Lemma 7.4.3. Any degree-two complex polynomial must have a complex root. Proof. This is a consequence of the quadratic formula and of the fact that any complex number has a square root. If P .x/ D ax 2 C bx C c, a ¤ 0, then the roots formally are x1 D

b C

p b 2  4ac ; 2a

x2 D

b 

p b 2  4ac : 2a

From DeMoivre’s theorem every complex number has a square root, hence x1 ; x2 exist in C. They of course are the same if b 2  4ac D 0. To go further we need the concept of the conjugate of a polynomial and some straightforward consequences of this idea. Definition 7.4.4. If P .x/ D a0 C    C an x n is a complex polynomial then its conjugate is the polynomial P .x/ D a0 C    C an x n . That is, the conjugate is the polynomial whose coefficients are the complex conjugates of those of P .x/. Lemma 7.4.5. For any P .x/ 2 CŒx we have: (1) P .z/ D P .z/ if z 2 C. (2) P .x/ is a real polynomial if and only if P .x/ D P .x/. (3) If P .x/Q.x/ D H.x/ then H .x/ D .P .x//.Q.x//.

Section 7.4 The Fundamental Theorem of Algebra

107

Proof. (1) Suppose z 2 C and P .z/ D a0 C    C an z n . Then P .z/ D a0 C    C an z n D a0 C a1 z C    C an z n D P .z/: (2) Suppose P .x/ is real then ai D ai for all its coefficients and hence P .x/ D P .x/. Conversely suppose P .x/ D P .x/. Then ai D ai for all its coefficients and hence ai 2 R for each ai and so P .x/ is a real polynomial. (3) The proof is a computation and left to the exercises. Lemma 7.4.6. Suppose G.x/ 2 CŒx. Then H.x/ D G.x/G.x/ 2 RŒx. Proof. H .x/ D G.x/G.x/ D G.x/G.x/ D G.x/G.x/ D G.x/G.x/ D H.x/. Therefore, H.x/ is a real polynomial. Lemma 7.4.7. If every nonconstant real polynomial has a complex root, then every nonconstant complex polynomial has a complex root. Proof. Let P .x/ 2 CŒx and suppose that every nonconstant real polynomial has at least one complex root. Let H.x/ D P .x/P .x/. From Lemma 7.4.6, H.x/ 2 RŒx. By supposition there exists a z0 2 C with H.z0 / D 0. Then P .z0 /P .z0 / D 0, and since C is a field it has no zero divisors. Hence either P .z0 / D 0 or P .z0 / D 0. In the first case z0 is a root of P .x/. In the second case P .z0 / D 0. Then from Lemma 7.4.5 P .z0 / D P .z0 / D P .z0 / D 0. Therefore, z0 is a root of P .x/. Now we come to the crucial lemma. Lemma 7.4.8. Any nonconstant real polynomial has a complex root. Proof. Let f .x/ D a0 C a1 x C    C an x n 2 RŒx with n  1, an ¤ 0. The proof is an induction on the degree n of f .x/. Suppose n D 2m q where q is odd. We do the induction on m. If m D 0 then f .x/ has odd degree and the theorem is true from Lemma 7.4.2. Assume then that the theorem is true for all degrees d D 2k q 0 where k < m and q 0 is odd. Now assume that the degree of f .x/ is n D 2m q. Suppose K 0 is the splitting field for f .x/ over R in which the roots are ˛1 ; : : : ; ˛n . We show that at least one of these roots must be in C. (In fact, all are in C but to prove the lemma we need only show at least one.) Let h 2 Z and form the polynomial H.x/ D

Y i<j

.x  .˛i C ˛j C h˛i ˛j //:

108

Chapter 7 Kronecker’s Theorem and Algebraic Closures

This is in K 0 Œx. In forming H.x/ we chose pairs of roots ¹˛i ; ˛j º, so the number of such pairs is the number of ways of choosing two elements out of n D 2m q elements. This is given by .2m q/.2m q  1/ D 2m1 q.2m q  1/ D 2m1 q 0 2 with q 0 odd. Therefore, the degree of H.x/ is 2m1 q 0 . H.x/ is a symmetric polynomial in the roots ˛1 ; : : : ; ˛n . Since ˛1 ; : : : ; ˛n are the roots of a real polynomial, from Lemma 7.3.14 any polynomial in the splitting field symmetric in these roots must be a real polynomial. Therefore, H.x/ 2 RŒx with degree 2m1 q 0 . By the inductive hypothesis, then, H.x/ must have a complex root. This implies that there exists a pair ¹˛i ; ˛j º with ˛i C ˛j C h˛i ˛j 2 C: Since h was an arbitrary integer, for any integer h1 there must exist such a pair ¹˛i ; ˛j º with ˛i C ˛j C h1 ˛i ˛j 2 C: Now let h1 vary over the integers. Since there are only finitely many such pairs ¹˛i ; ˛j º, it follows that there must be at least two different integers h1 ; h2 such that z1 D ˛i C ˛j C h1 ˛i ˛j 2 C

and

z2 D ˛i C ˛j C h2 ˛i ˛j 2 C:

Then z1  z2 D .h1  h2 /˛i ˛j 2 C and since h1 ; h2 2 Z  C it follows that ˛ ˛i j 2 C. But then h1 ˛i ˛j 2 C, from which it follows that ˛i C ˛j 2 C. Then, p.x/ D .x  ˛i /.x  ˛j / D x 2  .˛i C ˛j /x C ˛i ˛j 2 CŒx: However, p.x/ is then a degree-two complex polynomial and so from Lemma 7.4.3 its roots are complex. Therefore, ˛i ; ˛j 2 C, and therefore f .x/ has a complex root. It is now easy to give a proof of the fundamental theorem of algebra. From Lemma 7.4.8 every nonconstant real polynomial has a complex root. From Lemma 7.4.7 if every nonconstant real polynomial has a complex root, then every nonconstant complex polynomial has a complex root proving the fundamental theorem. Theorem 7.4.9. If E is a finite dimensional field extension of C, then E D C. Proof. Let a 2 E. Regard the elements 1; a; a2 ; : : : . These elements become linearly dependent over C and we get a nonconstant polynomial over C with root a. By the fundamental theorem of algebra we know that a 2 C. Corollary 7.4.10. If E is a finite dimensional field extension of R, then E D R or E D C.

Section 7.5 The Fundamental Theorem of Symmetric Polynomials

7.5

109

The Fundamental Theorem of Symmetric Polynomials

In the proof of the fundamental theorem of algebra that was given in the previous section we used the fact that any symmetric polynomial in n indeterminates is a polynomial in the elementary symmetric polynomials in these indeterminates. In this section we give a proof of this theorem. Let R be an integral domain with x1 ; : : : ; xn (independent) indeterminates over R and let RŒx1 ; : : : ; xn  be the polynomial ring in these indeterminates. Any polynomial f .x1 ; : : : ; xn / 2 RŒx1 ; : : : ; xn  is composed of a sum of pieces of the form ax1i1 : : : xnin with a 2 R. We first put an order on these pieces of a polynomial. j j The piece ax1i1 : : : xnin with a ¤ 0 is called higher than the piece bx11 : : : xnn with b ¤ 0 if the first one of the differences i1  j1 ; i2  j2 ; : : : ; in  jn that differs from zero is in fact positive. The highest piece of a polynomial f .x1 ; : : : ; xn / is denoted by HG.f /. Lemma 7.5.1. For f .x1 ; : : : ; xn /; g.x1 ; : : : ; xn / 2 RŒx1 ; : : : ; xn  we have HG.fg/ D HG.f / HG.g/: Proof. We use an induction on n, the number of indeterminates. It is clearly true for n D 1, and now assume that the statement holds for all polynomials in k indeterminates with k < n and n  2. Order the polynomials via exponents on the first indeterminate x1 so that f .x1 ; : : : ; xn / D x1r r .x2 ; : : : ; xn / C x1r1 r1 .x2 ; : : : ; xn / C    C 0 .x2 ; : : : ; xn / g.x1 ; : : : ; xn / D x1s

s .x2 ; : : : ; xn /

C  C Then HG.fg/ D x1rCs HG.r

s /.

HG.r

C x1s1

s1 .x2 ; : : : ; xn /

0 .x2 ; : : : ; xn /:

By the inductive hypothesis s/

D HG.r / HG.

s /:

Hence HG.fg/ D x1rCs HG.r / HG.

s/

D .x1r HG.r //.x1s HG.

s //

D HG.f / HG.g/:

110

Chapter 7 Kronecker’s Theorem and Algebraic Closures

The elementary symmetric polynomials in n indeterminates x1 ; : : : ; xn are s1 D x1 C x2 C    C xn s2 D x1 x2 C x1 x3 C    C xn1 xn s3 D x1 x2 x3 C x1 x2 x4 C    C xn2 xn1 xn :: : sn D x1    xn : These were found by forming the polynomial p.x; x1 ; : : : ; xn / D .x  x1 /    .x  xn /. The i -th elementary symmetric polynomial si in x1 ; : : : ; xn is then .1/i ai , where ai is the coefficient of x ni in p.x; x1 ; : : : ; xn /. In general, X sk D xi1 xi2    xik ; i1 0. Now g m 2 H , g t 2 H so g qt 2 H for any q since H is a subgroup. It follows that g m g qt D gmqt 2 H . This implies that gr 2 H . However this is a contradiction since r < t and t is the least positive power in H . It follows that r D 0 so m D qt . This implies that g m D gqt D .g t /q , that is gm is a multiple of g t . Therefore every element of H is a multiple of g t and therefore g t generates H and hence H is cyclic. Now suppose that d jn so that n D kd . Let H D hgk i, that is the subgroup of G generated by g k . We claim that H has order d and that any other subgroup H1 of G with order d coincides with H . Now .g k /d D g kd D g n D 1 so the order of gk divides d and hence is  d . Suppose that .g k /d1 D g kd1 D 1 with d1 < d . Then since the order of g is n we have n D kd jkd1 with d1 < d which is impossible. Therefore the order of g k is d and h D hgk i is a subgroup of G of order d . Now let H1 be a subgroup of G of order d . We must show that H1 D H . Let h 2 H1 so h D g t and hence g td D 1. It follows that njt d and so kd jt d and hence kjt , that is t D qk for some positive integer q. Therefore g t D .gk /q 2 H . Therefore H1  H and since they are of the same size H D H1 . Theorem 9.5.7. Let G D hgi be an infinite cyclic group. Then a subgroup H is of the form H D hg t i for a positive integer t . Further if t1 ; t2 are positive integers with t1 ¤ t2 then hg t1 i and hg t2 i are distinct. Proof. Let G D hgi be an infinite cyclic group and H a subgroup of G. As in the proof of Theorem 9.5.6 H must contain positive powers of the generator g. Let t be the smallest positive power of g such that g t 2 H . We claim that H D hgt i the cyclic subgroup of G generated by g t . Let h 2 H then h D gm for some positive integer m  t . Divide m by t to get m D qt C r

where r D 0 or 0 < r < t:

If r ¤ 0 then r D m  qt > 0. Now g m 2 H , g t 2 H so g qt 2 H for any q since H is a subgroup. It follows that g m g qt D gmqt 2 H . This implies that gr 2 H . However this is a contradiction since r < t and t is the least positive power in H . It follows that r D 0 so m D qt . This implies that g m D gqt D .g t /q , that is gm is a multiple of g t . Therefore every element of H is a multiple of g t and therefore g t generates H and hence H D hg t i. From the proof above in the subgroup hg t i the integer t is the smallest positive power of g in hg t i. Therefore if t1 ; t2 are positive integers with t1 ¤ t2 then hg t1 i and hg t2 i are distinct.

136

Chapter 9 Groups, Subgroups and Examples

Theorem 9.5.8. Let G D hgi be a cyclic group. Then (a) If G D hgi is finite of order n then g k is also a generator if and only if .k; n/ D 1. That is the generators of G are precisely those powers gk where k is relatively prime to n. (b) If G D hgi is infinite then the only generators are g; g1 . Proof. (a) Let G D hgi be a finite cyclic group of order n and suppose that .k; n/ D 1. Then there exist integers x; y with kx C ny D 1. It follows that g D g kxCny D .gk /x .gn /y D .g k /x since gn D 1. Hence g is a power of g k that implies every element of G is also a power of g k . Therefore g k is also a generator. Conversely suppose that gk is also a generator. Then g is a power of gk so there exists an x such that g D g kx . It follows that kx  1 modulo n and so there exists a y such that kx C ny D 1: This then implies that .k; n/ D 1. (b) If G D hgi is infinite then any power of g other than g 1 generates a proper subgroup. If g is a power of g n for some n so that g D g nx it follows that g nx1 D 1 so that g has finite order contradicting that G is infinite cyclic. Recall that for positive integers n the Euler phi-function is defined as follows. Definition 9.5.9. For any n > 0, let .n/ D number of integers less than or equal to n and relatively prime to n: Example 9.5.10. .6/ D 2 since among 1; 2; 3; 4; 5; 6 only 1; 5 are relatively prime to 6. Corollary 9.5.11. If G D hgi is finite of order n then there are .n/ generators for G where  is the Euler phi-function. Proof. From Theorem 9.5.8 the generators of G are precisely the powers gk where .k; n/ D 1. The numbers relatively prime to n are counted by the Euler phi-function. Recall that in an arbitrary group G, if g 2 G, then the order of g, denoted o.g/, is the order of the cyclic subgroup generated by g. Given two elements g; h 2 G in general there is no relationship between o.g/; o.h/ and the order of the product gh. However if they commute there is a very direct relationship.

Section 9.5 Generators and Cyclic Groups

137

Lemma 9.5.12. Let G be an arbitrary group and g; h 2 G both of finite order o.g/; o.h/. If g and h commute, that is gh D hg, then o.gh/ divides lcm.o.g/; o.h//. In particular if G is an abelian group then o.gh/j lcm.o.g/; o.h// for all g; h 2 G of finite order. Further if hgi \ hhi D ¹1º then o.gh/ D lcm.o.g/; o.h//. Proof. Suppose o.g/ D n and o.h/ D m are finite. If g; h commute then for any k we have .gh/k D gk hk . Let t D lcm.n; m/ then t D k1 m; t D k2 n. Hence .gh/t D g t ht D .g m /k1 .hn /k2 D 1: Therefore the order of gh is finite and divides t . Suppose that hgi \ hhi D ¹1º that is the cyclic subgroup generated by g intersects trivially with the cyclic subgroup generated by h. Let k D o.gh/ which we know is finite from the first part of the lemma. Let t D lcm.n; m/. We then have .gh/k D gk hk D 1 which implies that g k D hk . Since the cyclic subgroups have only trivial intersection this implies that g k D 1 and hk D 1. But then njk and mjk and hence t jk. Since kjt it follows that k D t. Recall that if m and n are relatively prime then lcm.m; n/ D mn. Further if the orders of g and h are relatively prime it follows from Lagrange’s theorem that hgi \ hhi D ¹1º. We then get the following. Corollary 9.5.13. If g; h commute and o.g/ and o.h/ are finite and relatively prime then o.gh/ D o.g/o.h/. Definition 9.5.14. If G is a finite abelian group then the exponent of G is the lcm of the orders of all elements of G. That is exp.G/ D lcm¹o.g/ W g 2 Gº: As a consequence of Lemma 9.5.12 we obtain Lemma 9.5.15. Let G be a finite abelian group. Then G contains an element of order exp.G/. e Proof. Suppose that exp.G/ D p1e1    pkk with pi distinct primes. By the definition of exp.G/ there is a gi 2 G with o.gi / D piei ri with pi and ri relatively prime. Let hi D giri . Then from Lemma 9.5.12 we get o.hi / D piei . Now let g D h1 h2    hk . From the corollary to Lemma 9.5.12 we have o.g/ D p1e1    pkek D exp.G/.

If K is a field then the multiplicative subgroup of nonzero elements of K is an abelian group K ? . The above results lead to the fact that a finite subgroup of K ? must actually be cyclic. Theorem 9.5.16. Let K be a field. Then any finite subgroup of K ? is cyclic.

138

Chapter 9 Groups, Subgroups and Examples

Proof. Let A  K ? with jAj D n. Suppose that m D exp.A/. Consider the polynomial f .x/ D x m  1 2 KŒx. Since the order of each element in A divides m it follows that am D 1 for all a 2 A and hence each a 2 A is a zero of the polynomial f .x/. Hence f .x/ has at least n zeros. Since a polynomial of degree m over a field can have at most m zeros it follows that n < m. From Lemma 9.5.15 there is an element a 2 A with o.a/ D m. Since jAj D n it follows that mjn and hence m < n. Therefore m D n and hence A D hai showing that A is cyclic. We close this section with two other results concerning cyclic groups. The first proves, using group theory, a very interesting number theoretic result concerning the Euler phi-function. Theorem 9.5.17. For n > 1 and for d  1 X .d / D n: d jn

Proof. Consider a cyclic group G of order n. For each d jn, d  1 there is a unique cyclic subgroup H of order d . H then has .d / generators. Each element in G generates its own cyclic subgroup H1 , say of order d and hence must be included in the .d / generators of H1 . Therefore X .d / D sum of the numbers of generators of the cyclic subgroups of G: d jn

But this must be the whole group and hence this sum is n. We shall make use of the above theorem directly in the following theorem. Theorem 9.5.18. If jGj D n and if for each positive d such that d jn, G has at most one cyclic subgroup of order d , then G is cyclic (and consequently, has exactly one cyclic subgroup of order d ). Proof. For each d jn, d > 0, let .d / D the number of elements of G of order d . Then X .d / D n: d jn

Now suppose that .d / ¤ 0 for a given d jn. Then there exists an a 2 G of order d which generates a cyclic subgroup, hai, of order d of G. We claim that all elements of G of order d are in hai. Indeed, if b 2 G with o.b/ D d and b … hai, then hbi is a second cyclic subgroup of order d , distinct from hai. This contradicts the hypothesis, so the claim is proved. Thus, if .d / ¤ 0, then P .d / D .d /. P In general, we have .d /  .d /, for all positive d jn. But n D d jn .d /  d jn .d /, by the previous theorem. It follows, clearly, from this that .d / D .d / for all d jn. In

Section 9.6 Exercises

139

particular, .n/ D .n/  1. Hence, there exists at least one element of G of order n; hence G is cyclic. This completes the proof. Corollary 9.5.19. If in a group G of order n, for each d jn, the equation x d D 1 has at most d solutions in G, then G is cyclic. Proof. The hypothesis clearly implies that G can have at most one cyclic subgroup of order d since all elements of such a subgroup satisfy the equation. So Theorem 9.5.17 applies to give our result. If H is a subgroup of a group G then G operates as a group of permutations on the set ¹aH W a 2 Rº of left cosets of H in G where R is a left transversal of H in G. This we can use to show that a finitely generated group has only finitely many subgroups of a given finite index. Theorem 9.5.20. Let G be a finitely generated group. The number of subgroups of index n < 1 is finite. Proof. Let H be a subgroup of index n. We choose a left transversal ¹c1 ; : : : ; cn º for H in G where c1 D 1 represents H . G permutes the set of cosets ci H by multiplication from the left. This induces a homomorphism H from G to Sn as follows. For each g 2 G let H .g/ be the permutation which maps i to j if gci H D cj H . H .g/ fixes the number 1 if and only if g 2 H because c1 H D H . Now, let H and L be two different subgroups of index n in G. Then there exists g 2 H with g … L and H .g/ ¤ L .g/, and hence H and L are different. Since G is finitely generated there are only finitely many homomorphisms from G to Sn . Therefore the number of subgroups of index n < 1 is finite.

9.6

Exercises

1.

Prove Lemma 9.1.4.

2.

Suppose that g 2 G and g m D 1 for some positive integer m. Let n be the smallest positive integer such that gn D 1. Show the set of elements ¹1; g; g2 ; : : : ; g n1 º are all distinct but for any other power g k we have g k D g t for some k D 0; 1; : : : ; n  1.

3.

Let G be a group and U1 ; U2 be finite subgroups of G. If jU1 j and jU2 j are relatively prime, then U1 \ U2 D ¹eº.

4.

Let A; B be subgroups of a finite group G. If jAj  jBj > jGj then A \ B ¤ ¹eº.

140 5.

Chapter 9 Groups, Subgroups and Examples

Let G be the set of all real matrices of the form Show:

 a b  b a

, where a2 C b 2 ¤ 0.

(a) G is a group. (b) For each n 2 N there is at least one element of order n in G. 6.

Let p be a prime, and let G D SL.2; p/. Show: (a) G has at least 2p  2 elements of order p (their exact number is p 2  1). (b) If p is odd, then x p D 1 for all x 2 G.

7.

Let p be a prime and a 2 Z. Show that ap  a mod p.

8.

Let F be a field. Show that the set of n  n matrices of determinant 1 over F forms a group.

9.

Here we outline a proof that every planar Euclidean congruence motion is either a rotation, translation, reflection or glide reflection. An isometry in this problem is a planar Euclidean congruence motion. Show: (a) If T is an isometry then it is completely determined by its action on a triangle – equivalent to showing that if T fixes three noncollinear points then it must be the identity. (b) If an isometry T has exactly one fixed point then it must be a rotation with that point as center. (c) If an isometry T has two fixed points then it fixes the line joining them. Then show that if T is not the identity it must be a reflection through this line. (d) If an isometry T has no fixed point but preserves orientation then it must be a translation. (e) If an isometry T has no fixed point but reverses orientation then it must be a glide reflection.

10. Let Pn be a regular n-gon and DN its group of symmetries. Show that jDn j D 2n. (Hint: First show that jDn j  2n and then exhibit 2n distinct symmetries.) 11. If A; B have the same cardinality, then there exists a bijection W A ! B. Define a map F W SA ! SB in the following manner: if f 2 SA , let F .f / be the permutation on B given by F .f /.b/ D .f . 1 .b///. Show that F is an isomorphism. 12. Prove Lemma 9.3.3.

Chapter 10

Normal Subgroups, Factor Groups and Direct Products

10.1

Normal Subgroups and Factor Groups

In rings we saw that there were certain special types of subrings, called ideals, that allowed us to define factor rings. The analogous object for groups is called a normal subgroup which we will define and investigate in this section. Definition 10.1.1. Let G be an arbitrary group and suppose that H1 and H2 are subgroups of G. We say that H2 is conjugate to H1 if there exists an element a 2 G such that H2 D aH1 a1 . H1 ; H2 are the called conjugate subgroups of G. Lemma 10.1.2. Let G be an arbitrary group. Then the relation of conjugacy is an equivalence relation on the set of subgroups of G. Proof. We must show that conjugacy is reflexive, symmetric and transitive. If H is a subgroup of G then 11 H1 D H and hence H is conjugate to itself and therefore the relation is reflexive. Suppose that H1 is conjugate to H2 . Then there exists a g 2 G with g 1 H1 g D H2 . This implies that gH2 g 1 D H1 . However .g 1 /1 D g and hence letting g 1 D g1 we have g11 H2 g1 D H1 : Therefore H2 is conjugate to H1 and conjugacy is symmetric. Finally suppose that H1 is conjugate to H2 and H2 is conjugate to H3 . Then there exist g1 ; g2 2 G with H2 D g11 H1 g1 and H3 D g21 H2 g2 . Then H3 D g21 g11 H1 g1 g2 D .g1 g2 /1 H1 .g1 g2 /: Therefore H3 is conjugate to H2 and conjugacy is transitive. Lemma 10.1.3. Let G be an arbitrary group. Then for g 2 G the map g W a ! g 1 ag is an automorphism on G. Proof. For a fixed g 2 G define the map f W G ! G by f .a/ D g 1 ag for a 2 G. We must show that this is a homomorphism and that it is one-to-one and onto.

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Let a1 ; a2 2 G. Then f .a1 a2 / D g 1 a1 a2 g D .g1 a1 g/.g 1 a2 g/ D f .a1 /f .a2 /: Hence f is a homomorphism. If f .a1 / D f .a2 / then g 1 a1 g D g 1 a2 g. Clearly by the cancellation law we then have a1 D a2 and hence f is one-to-one. Finally let a 2 G and let a1 D gag 1 . Then a D g 1 a1 g and hence f .a1 / D a. It follows that f is onto and therefore f is an automorphism on G. In general a subgroup H of a group G may have many different conjugates. However in certain situations the only conjugate of a subgroup H is H itself. If this is the case we say that H is a normal subgroup. We will see shortly that this is precisely the analog for groups of the concept of an ideal in rings. Definition 10.1.4. Let G be an arbitrary group. A subgroup H is a normal subgroup of G, which we denote by H G G, if g 1 Hg D H for all g 2 G. Since the conjugation map is an isomorphism it follows that if g 1 Hg  H then D H . Hence in order to show that a subgroup is normal we need only show g inclusion. 1 Hg

Lemma 10.1.5. Let N be a subgroup of a group G. Then if aNa1  N for all a 2 G, then aNa1 D N . In particular, aNa1  N for all a 2 G implies that N is a normal subgroup. Notice that if g 1 Hg D H then Hg D gH . That is as sets the left coset gH is equal to the right coset Hg. Hence for each h1 2 H there is an h2 2 H with gh1 D h2 g. If H G G this is true for all g 2 G. Further if H is normal then for the product of two cosets g1 H and g2 H we have .g1 H /.g2 H / D g1 .Hg2 /H D g1 g2 .HH / D g1 g2 H: If .g1 H /.g2 H / D .g1 g2 /H for all g1 ; g2 2 G we necessarily have gHg1 D H for all g 2 G. Hence we have proved: Lemma 10.1.6. Let H be a subgroup of a group G. Then the following are equivalent: (1) H is a normal subgroup of G. (2) g 1 Hg D H for all g 2 G. (3) gH D Hg for all g 2 G. (4) .g1 H /.g2 H / D .g1 g2 /H for all g1 ; g2 2 G.

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This is precisely the condition needed to construct factor groups. First we give some examples of normal subgroups. Lemma 10.1.7. Every subgroup of an abelian group is normal. Proof. Let G be abelian and H a subgroup of G. Suppose g 2 G then gh D hg for all h 2 H since G is abelian. It follows that gH D Hg. Since this is true for every g 2 G it follows that H is normal. Lemma 10.1.8. Let H  G be a subgroup of index 2, that is ŒG W H  D 2. Then H is normal in G. Proof. Suppose that ŒG W H  D 2. We must show that gH D Hg for all g 2 G. If g 2 H then clearly H D gH D Hg. Therefore we may assume that g is not in H . Then there are only 2 left cosets and 2 right cosets. That is, G D H [ gH D H [ Hg: Since the union is a disjoint union we must have gH D Hg and hence H is normal.

Lemma 10.1.9. Let K be any field. Then the group SL.n; K/ is a normal subgroup of GL.n; K/ for any positive integer n. Proof. Recall that GL.n; K/ is the group of n  n matrices over the field K with nonzero determinant while SL.n; K/ is the subgroup of n  n matrices over the field K with determinant equal to 1. Let U 2 SL.n; K/ and T 2 GL.n; K/. Consider T 1 U T . Then det.T 1 U T / D det.T 1 / det.U / det.T / D det.U / det.T 1 T / D det.U / det.I / D det.U / D 1: Hence T 1 U T 2 SL.n; K/ for any U 2 SL.n; K/ and any T 2 GL.n; K/. It follows that T 1 SL.n; K/T  SL.n; K/ and therefore SL.n; K/ is normal in GL.n; K/. The intersection of normal subgroups is again normal and the product of normal subgroups is normal. Lemma 10.1.10. Let N1 ; N2 be normal subgroups of the group G. Then (1) N1 \ N2 is a normal subgroup of G. (2) N1 N2 is a normal subgroup of G. (3) If H is any subgroup of G then N1 \H is a normal subgroup of H and N1 H D HN1 .

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Proof. (a) Let n 2 N1 \ N2 and g 2 G. Then g 1 ng 2 N1 since N1 is normal. Similarly g 1 ng 2 N2 since N2 is normal. Hence g1 ng 2 N1 \ N2 . It follows that g 1 .N1 \ N2 /g  N1 \ N2 and therefore N1 \ N2 is normal. (b) Let n1 2 N1 ; n2 2 N2 . Since N1 ; N2 are both normal N1 N2 D N2 N1 as sets and the complex N1 N2 forms a subgroup of G. Let g 2 G and n1 n2 2 N1 N2 . Then g 1 .n1 n2 /g D .g 1 n1 g/.g 1 n2 g/ 2 N1 N2 since g 1 n1 g 2 N1 and g 1 n2 g 2 N2 . Therefore N1 N2 is normal in G. (c) Let h 2 H and n 2 N \ H . Then as in part (a) h1 nh 2 N \ H and therefore N \ H is a normal subgroup of H . The same argument as in part (b) shows that N1 H D HN1 . We now construct factor groups or quotient groups of a group modulo a normal subgroup. Definition 10.1.11. Let G be an arbitrary group and H a normal subgroup of G. Let G=H denote the set of distinct left (and hence also right) cosets of H in G. On G=H define the multiplication .g1 H /.g2 H / D g1 g2 H for any elements g1 H; g2 H in G=H . Theorem 10.1.12. Let G be a group and H a normal subgroup of G. Then G=H under the operation defined above forms a group. This group is called the factor group or quotient group of G modulo H . The identity element is the coset 1H D H and the inverse of a coset gH is g 1 H . Proof. We first show that the operation on G=N is well-defined. Suppose that a0 N D aN and b 0 N D bN , then b 0 2 bN and so b 0 D bn1 . Similarly a0 D an2 where n1 ; n2 2 N . Therefore a0 b 0 N D an2 bn1 N D an2 bN since n1 2 N . But b 1 n2 b D n3 2 N , since N is normal, so the right-hand side of the equation can be written as an2 bN D abN: Thus we have shown that if N G G then a0 b 0 N D abN , and the operation on G=N is indeed, well-defined. The associative law is true because coset multiplication as defined above uses the ordinary group operation which is by definition associative. The coset N serves as the identity element of G=N . Notice that aN  N D aN 2 D aN

Section 10.1 Normal Subgroups and Factor Groups

145

and N  aN D aN 2 D aN: The inverse of aN is a1 N since aNa1 N D aa1 N 2 D N: We emphasize that the elements of G=N are cosets and thus subsets of G. If jGj < 1, then jG=N j D ŒG W N , the member of cosets of N in G. It is also to be emphasized that in order for G=N to be a group N must be a normal subgroup of G. In some cases properties of G are preserved in factor groups. Lemma 10.1.13. If G is abelian then any factor group of G is also abelian. If G is cyclic then any factor group of G is also cyclic. Proof. Suppose that G is abelian and H is a subgroup of G. H is necessarily normal from Lemma 10.1.7 so that we can form the factor group G=H . Let g1 H; g2 H 2 G=H . Since G is abelian we have g1 g2 D g2 g1 . Then in G=H , .g1 H /.g2 H / D .g1 g2 /H D .g2 g1 /H D .g2 H /.g1 H /: Therefore G=H is abelian. We leave the proof of the second part to the exercises. An extremely important concept is when a group contains no proper normal subgroups other than the identity subgroup ¹1º. Definition 10.1.14. A group G ¤ ¹1º is simple provided that N G G implies N D G or N D ¹1º. One of the most outstanding problems in group theory has been to give a complete classification of all finite simple groups. In other words, this is the program to discover all finite simple groups and to prove that there are no more to be found. This was accomplished through the efforts of many mathematicians. The proof of this magnificent result took thousands of pages. We refer the reader to [18] for a complete discussion of this. We give one elementary example. Lemma 10.1.15. Any finite group of prime order is simple and cyclic. Proof. Suppose that G is a finite group and jGj D p where p is a prime. Let g 2 G with g ¤ 1. Then hgi is a nontrivial subgroup of G so its order divides the order of G by Lagrange’s theorem. Since g ¤ 1 and p is a prime we must have jhgij D p. Therefore hgi is all of G, that is G D hgi and hence G is cyclic. The argument above shows that G has no nontrivial proper subgroups and therefore no nontrivial normal subgroups. Therefore G is simple. In the next chapter we will examine certain other finite simple groups.

146

10.2

Chapter 10 Normal Subgroups, Factor Groups and Direct Products

The Group Isomorphism Theorems

In Chapter 1 we saw that there was a close relationship between ring homomorphisms and factor rings. In particular to each ideal, and consequently to each factor ring, there is a ring homomorphism that has that ideal as its kernel. Conversely to each ring homomorphism its kernel is an ideal and the corresponding factor ring is isomorphic to the image of the homomorphism. This was formalized in Theorem 1.5.7 which we called the ring isomorphism theorem. We now look at the group theoretical analog of this result, called the group isomorphism theorem. We will then examine some consequences of this result that will be crucial in the Galois theory of fields. Definition 10.2.1. If G1 and G2 are groups and f W G1 ! G2 is a group homomorphism then the kernel of f , denoted ker.f /, is defined as ker.f / D ¹g 2 G1 W f .g/ D 1º: That is the kernel is the set of the elements of G1 that map onto the identity of G2 . The image of f , denoted im.f /, is the set of elements of G2 mapped onto by f from elements of G1 . That is im.f / D ¹g 2 G2 W f .g1 / D g2 for some g1 2 G1 º: Note that if f is a surjection then im.f / D G2 . As with ring homomorphisms the kernel measures how far a homomorphism is from being an injection, that is, a one-to-one mapping. Lemma 10.2.2. Let G1 and G2 are groups and f W G1 ! G2 a group homomorphism. Then f is injective if and only if ker.f / D ¹1º. Proof. Suppose that f is injective. Since f .1/ D 1 we always have 1 2 ker.f /. Suppose that g 2 ker.f /. Then f .g/ D f .1/. Since f is injective this implies that g D 1 and hence ker.f / D ¹1º. Conversely suppose that ker.f / D ¹1º and f .g1 / D f .g2 /. Then f .g1 /.f .g2 //1 D 1 H) f .g1 g21 / D 1 H) g1 g21 2 ker.f /: Then since ker.f / D ¹1º we have g1 g21 D 1 and hence g1 D g2 . Therefore f is injective. We now state the group isomorphism theorem. This is entirely analogous to the ring isomorphism theorem replacing ideals by normal subgroups. We note that this theorem is sometimes called the first group isomorphism theorem.

Section 10.2 The Group Isomorphism Theorems

147

Theorem 10.2.3 (group isomorphism theorem). (a) Let G1 and G2 be groups and f W G1 ! G2 a group homomorphism. Then ker.f / is a normal subgroup of G1 , im.f / is a subgroup of G2 and G= ker.f / Š im.f /: (b) Conversely suppose that N is a normal subgroup of a group G. Then there exists a group H and a homomorphism f W G ! H such that ker.f / D N and im.f / D H . Proof. (a) Since 1 2 ker.f / the kernel is nonempty. Suppose that g1 ; g2 2 ker.f /. Then f .g1 / D f .g2 / D 1. It follows that f .g1 g21 / D f .g1 /.f .g2 //1 D 1. Hence g1 g21 2 ker.f / and therefore ker.f / is a subgroup of G1 . Further for any g 2 G1 we have f .g1 g1 g/ D .f .g//1 f .g1 /f .g/ D .f .g//1  1  f .g/ D f .g 1 g/ D f .1/ D 1: Hence g 1 g1 g 2 ker.f / and ker.f / is a normal subgroup. It is straightforward to show that im.f / is a subgroup of G2 . Consider the map fO W G= ker.f / ! im.f / defined by fO.g ker.f // D f .g/: We show that this is an isomorphism. Suppose that g1 ker.f / D g2 ker.f / then g1 g21 2 ker.f / so that f .g1 g21 / D 1. This implies that f .g1 / D f .g2 / and hence the map fO is well-defined. Now fO.g1 ker.f /g2 ker.f // D fO.g1 g2 ker.f // D f .g1 g2 / D f .g1 /f .g2 / D fO.g1 ker.f //fO.g2 ker.f // and therefore fO is a homomorphism. Suppose that fO.g1 ker.f // D fO.g2 ker.f // then f .g1 / D f .g2 / and hence g1 ker.f / D g2 ker.f /. It follows that fO is injective. Finally suppose that h 2 im.f /. Then there exists a g 2 G1 with f .g/ D h. Then fO.g ker.f // D h and fO is a surjection onto im.f /. Therefore fO is an isomorphism completing the proof of part (a). (b) Conversely suppose that N is a normal subgroup of G. Define the map f W G ! G=N by f .g/ D gN for g 2 G. By the definition of the product in the quotient group G=N it is clear that f is a homomorphism with im.f / D G=N . If g 2 ker.f / then f .g/ D gN D N since N is the identity in G=N . However this implies that g 2 N and hence it follows that ker.f / D N completing the proof.

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There are two related theorems that are called the second isomorphism theorem and the third isomorphism theorem. Theorem 10.2.4 (second isomorphism theorem). Let N be a normal subgroup of a group G and U a subgroup of G. Then U \ N is normal in U and .UN /=N Š U=.U \ N /: Proof. From Lemma 10.1.10 we know that U \ N is normal in U . Define the map ˛ W UN ! U=U \ N by ˛.un/ D u.U \ N /. If un D u0 n0 then u01 u D n0 n1 2 U \ N . Therefore u0 .U \ N / D u.U \ N / and hence the map ˛ is well-defined. Suppose that un; u0 n0 2 UN Since N is normal in G we have that unu0 n0 2 uu0 N . Hence unu0 n0 D uu0 n00 with n00 2 N . Then ˛.unu0 n0 / D ˛.uu0 n/ D uu0 .U \ N /: However U \ N is normal in U so uu0 .U \ N / D u.U \ N /u0 .U \ N / D ˛.un/˛.u0 n0 /: Therefore ˛ is a homomorphism. We have im.˛/ D U=.U \ N / by definition. Suppose that un 2 ker.˛/. Then ˛.un/ D U \ N  N which implies u 2 N . Therefore ker.f / D N . From the group isomorphism theorem we then have UN=N Š U=.U \ N / proving the theorem. Theorem 10.2.5 (third isomorphism theorem). Let N and M be normal subgroups of a group G with N a subgroup of M . Then M=N is a normal subgroup in G=N and .G=N /=.M=N / Š G=M: Proof. Define the map ˇ W G=N ! G=M by ˇ.gN / D gM: It is straightforward that ˇ is well-defined and a homomorphism. If gN 2 ker.ˇ/ then ˇ.gN / D gM D M and hence g 2 M . It follows that ker.ˇ/ D M=N . In particular this shows that M=N is normal in G=N . From the group isomorphism theorem then .G=N /=.M=N / Š G=M:

Section 10.3 Direct Products of Groups

149

For a normal subgroup N in G the homomorphism f W G ! G=N provides a one-to-one correspondence between subgroups of G containing N and the subgroups of G=N . This correspondence will play a fundamental role in the study of subfields of a field. Theorem 10.2.6 (correspondence theorem). Let N be a normal subgroup of a group G and let f be the corresponding homomorphism f W G ! G=N . Then the mapping  W H ! f .H / where H is a subgroup of G containing N provides a one-to-one correspondence between all the subgroups of G=N and the subgroups of G containing N . Proof. We first show that the mapping  is surjective. Let H1 be a subgroup of G=N and let H D ¹g 2 G W f .g/ 2 H1 º: We show that H is a subgroup of G and that N  H . If g1 ; g2 2 H then f .g1 / 2 H1 and f .g2 / 2 H1 . Therefore f .g1 /f .g2 / 2 H1 and hence f .g1 g2 / 2 H1 . Therefore g1 g2 2 H . In an identical fashion g11 2 H . Therefore H is a subgroup of G. If n 2 N then f .n/ D 1 2 H1 and hence n 2 H . Therefore N  H showing that the map  is surjective. Suppose that .H1 / D .H2 / where H1 and H2 are subgroups of G containing N . This implies that f .H1 / D f .H2 /. Let g1 2 H1 . Then f .g1 / D f .g2 / for some g2 2 H2 . Then g1 g21 2 ker.f / D N  H2 . It follows that g1 g21 2 H2 so that g1 2 H2 . Hence H1  H2 . In a similar fashion H2  H1 and therefore H1 D H2 . It follows that  is injective.

10.3

Direct Products of Groups

In this section we look at a very important construction, the direct product, which allows us to build new groups out of existing groups. This construction is the analog for groups of the direct sum of rings. As an application of this construction, in the next section we present a theorem which completely describes the structure of finite abelian groups. Let G1 ; G2 be groups and let G be the Cartesian product of G1 and G2 . That is G D G1  G2 D ¹.a; b/ W a 2 G1 ; b 2 G2 º: On G define .a1 ; b1 /  .a2 ; b2 / D .a1 a2 ; b1 b2 /: With this operation it is direct to verify the groups axioms for G and hence G becomes a group.

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Theorem 10.3.1. Let G1 ; G2 be groups and G the Cartesian product G1  G2 with the operation defined above. Then G forms a group called the direct product of G1 and G2 . The identity element is .1; 1/ and .g; h/1 D .g1 ; h1 /. This construction can be iterated to any finite number of groups (also to an infinite number but we won’t consider that here) G1 ; : : : ; Gn to form the direct product G1  G2      Gn . Theorem 10.3.2. For groups G1 and G2 we have G1  G2 Š G2  G1 and G1  G2 is abelian if and only if each Gi , i D 1; 2, is abelian. Proof. The map .a; b/ ! .b; a/ where a 2 G1 ; b 2 G2 provides an isomorphism from G1  G2 ! G2  G1 . Suppose that both G1 ; G2 are abelian. Then if a1 ; a2 2 G1 ; b1 ; b2 2 G2 we have .a1 ; b1 /.a2 ; b2 / D .a1 a2 ; b1 b2 / D .a2 a1 ; b2 b1 / D .a2 ; b2 /.a1 ; b1 / and hence G1  G2 is abelian. Conversely suppose G1  G2 is abelian and suppose that a1 ; a2 2 G1 . Then for the identity 1 2 G2 we have .a1 a2 ; 1/ D .a1 ; 1/.a2 ; 1/ D .a2 ; 1/.a1 ; 1/ D .a2 a1 ; 1/: Therefore a1 a2 D a2 a1 and G1 is abelian. Identically G2 is abelian. We show next that in G1  G2 there are normal subgroups H1 ; H2 with H1 Š G1 and H2 Š G2 . Theorem 10.3.3. Let G D G1  G2 . Let H1 D ¹.a; 1/ W a 2 G1 º and H2 D ¹.1; b/ W b 2 G2 º. Then both H1 and H2 are normal subgroups of G with G D H1 H2 and H1 \ H2 D ¹1º. Further H1 Š G1 ; H2 Š G2 and G=H1 Š G2 and G=H2 Š G1 . Proof. Map G1  G2 onto G2 by .a; b/ ! b. It is clear that this map is a homomorphism and that the kernel is H1 D ¹.a; 1/ W a 2 G1 º. This establishes that H1 is a normal subgroup of G and that G=H1 Š G2 . In an identical fashion we get that G=H2 Š G1 . The map .a; 1/ ! a provides the isomorphism from H1 onto G1 . If the factors are finite it is easy to find the order of G1  G2 . The size of the Cartesian product is just the product of the sizes of the factors. Lemma 10.3.4. If jG1 j and jG2 j are finite then jG1  G2 j D jG1 jjG2 j. Now suppose that G is a group with normal subgroups G1 ; G2 such that G D G1 G2 and G1 \ G2 D ¹1º. Then we will show that G is isomorphic to the direct product G1  G2 . In this case we say that G is the internal direct product of its subgroups and that G1 ; G2 are direct factors of G.

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151

Theorem 10.3.5. Suppose that G is a group with normal subgroups G1 ; G2 such that G D G1 G2 and G1 \ G2 D ¹1º. Then G is isomorphic to the direct product G1  G2 . Proof. Since G D G1 G2 each element of G has the form ab with a 2 G1 ; b 2 G2 . We first show that each a 2 G1 commutes with each b 2 G2 . Consider the element aba 1 b 1 . Since G1 is normal ba1 b 1 2 G1 which implies that abab 1 2 G1 . Since G2 is normal aba1 2 G2 which implies that aba1 b 1 2 G2 . Therefore aba1 b 1 2 G1 \ G2 D ¹1º and hence aba1 b 1 D 1 so that ab D ba. Now map G onto G1  G2 by f .ab/ ! .a; b/. We claim that this is an isomorphism. It is clearly onto. Now f ..a1 b1 /.a2 b2 // D f .a1 a2 b1 b2 / D .a1 a2 ; b1 b2 / D .a1 ; b1 /.a2 ; b2 / D f ..a1 ; b1 //f .a2 ; b2 // so that f is a homomorphism. The kernel is G1 \ G2 D ¹1º and so f is an isomorphism. Although the end resulting groups are isomorphic we call G1  G2 an external direct product if we started with the groups G1 ; G2 and constructed G1  G2 and call G1  G2 an internal direct product if we started with a group G having normal subgroups as in the theorem.

10.4

Finite Abelian Groups

We now use the results of the last section to present a theorem that completely provides the structure of finite abelian groups. This theorem is a special case of a general result on modules that we will examine in detail later in the book. Theorem 10.4.1 (basis theorem for finite abelian groups). Let G be a finite abelian group. Then G is a direct product of cyclic groups of prime power order. Before giving the proof we give two examples showing how this theorem leads to the classification of finite abelian groups. Since all cyclic groups of order n are isomorphic to .Zn ; C/ we will denote a cyclic group of order n by Zn . Example 10.4.2. Classify all abelian groups of order 60. Let G be an abelian group of order 60. From Theorem 10.4.1 G must be a direct product of cyclic groups of prime power order. Now 60 D 22  3  5 so the only primes involved are 2, 3 and 5. Hence the cyclic group involved in the direct product decomposition of G have order either 2, 4, 3 or 5 (by Lagrange’s theorem they must be divisors of 60). Therefore G

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must be of the form G Š Z4  Z3  Z5 G Š Z2  Z2  Z3  Z5 : Hence up to isomorphism there are only two abelian groups of order 60. Example 10.4.3. Classify all abelian groups of order 180. Now 180 D 22  32  5 so the only primes involved are 2, 3 and 5. Hence the cyclic group involved in the direct product decomposition of G have order either 2, 4, 3, 9 or 5 (by Lagrange’s theorem they must be divisors of 180). Therefore G must be of the form G Š Z4  Z9  Z5 G Š Z2  Z2  Z9  Z5 G Š Z4  Z3  Z3  Z5 G Š Z2  Z2  Z3  Z3  Z5 : Hence up to isomorphism there are four abelian groups of order 180. The proof of Theorem 10.4.1 involves the following lemmas. Lemma 10.4.4. Let G be a finite abelian group and let pjjGj where p is a prime. Then all the elements of G whose orders are a power of p form a normal subgroup of G. This subgroup is called the p-primary component of G, which we will denote by Gp . Proof. Let p be a prime with pjjGj and let a and b be two elements of G of order a power of p. Since G is abelian the order of ab is the lcm of the orders which is again a power of p. Therefore ab 2 Gp . The order of a1 is the same as the order of a so a 1 2 Gp and therefore Gp is a subgroup. Lemma 10.4.5. Let G be a finite abelian group of order n. Suppose that n D e p1e1    pkk with p1 ; : : : ; pk distinct primes. Then G Š Gp1      Gpk where Gpi is the pi -primary component of G. Proof. Each Gpi is normal since G is abelian and since distinct primes are relatively prime the intersection of the Gpi is the identity. Therefore Lemma 10.4.5 will follow by showing that each element of G is a product of elements in the Gp1 .

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Section 10.4 Finite Abelian Groups

Let g 2 G. Then the order of g is p1f1    pkfk . We write this as pifi m with f

.m; pi / D 1. Then g m has order pi i and hence is in Gpi . Now since p1 ; : : : ; pk are relatively prime there exists m1 ; : : : ; mk with f

f

m1 p1 1 C    C mk pk k D 1 and hence

f1

fk

g D .g p1 /m1    .g pk /mk : Therefore g is a product of elements in the Gpi . We next need the concept of a basis. Let G be any finitely generated abelian group (finite or infinite) and let g1 ; : : : ; gn be a set of generators for G. The generators g1 ; : : : ; gn form a basis if G D hg1 i      hgn i; that is G is the direct product of the cyclic subgroups generated by the gi . The basis theorem for finite abelian groups says that any finite abelian group has a basis. Suppose that G is a finite abelian group with a basis g1 ; : : : ; gk so that G D hg1 i      hgk i. Since G is finite each gi has finite order say mi . It follows than from the fact that G is a direct product that each g 2 G can be expressed as g D g1n1    gknk and further the integers n1 ; : : : ; nk are unique modulo the order of gi . Hence each integer ni can be chosen in the range 0; 1; : : : ; mi  1 and within this range for the element g the integer ni is unique. From the previous lemma each finite abelian group splits into a direct product of its p-primary components for different primes p. Hence to complete the proof of the basis theorem we must show that any finite abelian group of order p m for some prime p has a basis. We call an abelian group of order p m an abelian p-group. Consider an abelian group G of order p m for a prime p. It is somewhat easier to complete the proof if we consider the group using additive notation. That is the operation is considered C, the identity as 0 and powers are given by multiples. Hence if an element g 2 G has order p k then in additive notation p k g D 0. A set of elements g1 ; : : : ; gk is then a basis for G if each g 2 G can be expressed uniquely as g D m1 g1 C    C mk gk where the mi are unique modulo the order of gi . We say that the g1 ; : : : ; gk are independent and this is equivalent to the fact that whenever m1 g1 C    C mk gk D 0 then mi  0 modulo the order of gi . We now prove that any abelian p-group has a basis. Lemma 10.4.6. Let G be a finite abelian group of prime power order p n for some prime p. Then G is a direct product of cyclic groups.

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Notice that in the group G we have p n g D 0 for all g 2 G as a consequence of Lagrange’s theorem. Further every element has order a power of p. The smallest power of p say p r such that p r g D 0 for all g 2 G is called the exponent of G. Any finite abelian p-group must have some exponent p r . Proof. The proof of this lemma is by induction on the exponent. The lowest possible exponent is p so suppose first that pg D 0 for all g 2 G. Since G is finite it has a finite system of generators. Let S D ¹g1 ; : : : ; gk º be a minimal set of generators for G. We claim that this is a basis. Since this is a set of generators to show its a basis we must show that they are independent. Hence suppose that we have m1 g1 C    C mk gk D 0

(1)

for some set of integers mi . Since the order of each gi is p, as explained above we may assume that 0  mi < p for i D 1; : : : ; k. Suppose that one mi ¤ 0. Then .mi ; p/ D 1 and hence there exists an xi with mi xi  1 mod p (see Chapter 4). Multiplying the equation (1) by xi we get modulo p, m1 xi g1 C    C gi C    C mk xi gk ; and rearranging gi D m1 xi g1      mk xk gk ; But then gi can be expressed in terms of the other gj and therefore the set ¹g1 ; : : : ; gk º is not minimal. It follows that g1 ; : : : ; gk constitute a basis and the lemma is true for the exponent p. Now suppose that any finite abelian group of exponent p n1 has a basis and assume that G has exponent p n . Consider the set G D pG D ¹pg W g 2 Gº. It is straightforward that this forms a subgroup (see exercises). Since p n g D 0 for all g 2 G it follows that p n1 g D 0 for all g 2 G and so the exponent of G  p n1 . By the inductive hypothesis G has a basis S D ¹pg1 ; : : : ; pgk º: Consider the set ¹g1 ; : : : ; gk º and adjoin to this set the set of all elements h 2 G satisfying ph D 0. Call this set S1 so that we have S1 D ¹g1 ; : : : ; gk ; h1 ; : : : ; h t º: We claim that S1 is a set of generators for G. Let g 2 G. Then pg 2 G which has the basis pg1 ; : : : ; pgk so that pg D m1 pg1 C    C mk pgk : This implies that p.g  m1 g1      mk gk / D 0

155

Section 10.4 Finite Abelian Groups

so that g1  m1 g1      mk gk must be one of the hi . Hence g  m 1 g 1      m k g k D hi

so that g D m1 g1 C    C mk gk C hi

proving the claim. Now S1 is finite so there is a minimal subset of S1 that is still a generating system for G. Call this S0 and suppose that S0 , renumbering if necessary, is S0 D ¹g1 ; : : : ; gr ; h1 ; : : : ; hs º with phi D 0 for i D 1; : : : ; s: The subgroup generated by h1 ; : : : ; hs has exponent p so by inductive hypothesis has a basis. We may assume than that h1 ; : : : ; hs is a basis for this subgroup and hence is independent. We claim now that g1 ; : : : ; gr ; h1 ; : : : ; hs are independent and hence form a basis for G. Suppose that m1 gr C    C mr gr C n1 h1 C    C ns hs D 0

(2)

for some integers m1 ; : : : ; mr ; h1 ; : : : ; hs . Each mi ; ni must be divisible by p. Suppose for example that some mi is not. Then .mi ; p/ D 1 and then .mi ; p n / D 1. This implies that there exists an xi with mi xi  1 mod p n . Multiplying through by xi and rearranging we then obtain gi D m1 xi g1      ns xi hs : Therefore gi can be expressed in terms of the remaining elements of S0 contradicting the minimality of S0 . An identical argument works if some ni is not divisible by p. Therefore the relation (2) takes the form a1 pg1 C    C ar pgr C b1 ph1 C    C bs phs D 0:

(3)

Each of the terms phi D 0 so that (3) becomes ap g1 C    C ar pgr D 0: The g1 ; : : : ; gr are independent and hence ai p D 0 for each i and hence ai D 0. Now (2) becomes n1 h1 C    C ns hs D 0: However h1 ; : : : ; hs are independent so each ni D 0 completing the claim. Therefore the whole group G has a basis proving the lemma by induction. For more details see the proof of the general result on modules over principal ideal domains later in the book. There is also an additional elementary proof for the basis theorem for finitely generated abelian groups.

156

10.5

Chapter 10 Normal Subgroups, Factor Groups and Direct Products

Some Properties of Finite Groups

Classification is an extremely important concept in algebra. A large part of the theory is devoted to classifying all structures of a given type, for example all UFD’s. In most cases this is not possible. Since for a given finite n there are only finitely many group tables it is theoretically possible to classify all groups of order n. However even for small n this becomes impractical. We close the chapter by looking at some further results on finite groups and then using these to classify all the finite groups up to order 10. Before stating the classification we give some further examples of groups that are needed. Example 10.5.1. In Example 9.2.6 we saw that the symmetry group of an equilateral triangle had 6 elements and is generated by elements r and f which satisfy the relations r 3 D f 2 D 1, f 1 rf D r 1 , where r is a rotation of 120ı about the center of the triangle and f is a reflection through an altitude. This was called the dihedral group D3 of order 6. This can be generalized to any regular n-gon. If D is a regular n-gon, then the symmetry group Dn has 2n elements and is called the dihedral group of order 2n. It is generated by elements r and f which satisfy the relations r n D f 2 D 1, f 1 rf D r n1 , where r is a rotation of 2 n about the center of the n-gon and f is a reflection. Hence, D4 , the symmetries of a square, has order 8 and D5 , the symmetries of a regular pentagon, has order 10. Example 10.5.2. Let i; j; k be the generators of the quaternions. Then we have i 2 D j 2 D k 2 D 1;

.1/2 D 1

and

ij k D 1:

These elements then form a group of order 8 called the quaternion group denoted by Q. Since ij k D 1 we have ij D j i , and the generators i and j satisfy the relations i 4 D j 4 D 1, i 2 D j 2 , ij D i 2 j i . We now state the main classification and then prove it in a series of lemmas. Theorem 10.5.3. Let G be a finite group. (a) If jGj D 2 then G Š Z2 . (b) If jGj D 3 then G Š Z3 . (c) If jGj D 4 then G Š Z4 or G Š Z2  Z2 . (d) If jGj D 5 then G Š Z5 . (e) If jGj D 6 then G Š Z6 Š Z2  Z3 or G Š D3 , the dihedral group with 6 elements. (Note D3 Š S3 the symmetric group on 3 symbols.) (f) If jGj D 7 then G Š Z7 .

Section 10.5 Some Properties of Finite Groups

157

(g) If jGj D 8 then G Š Z8 or G Š Z4  Z2 or G Š Z2  Z2  Z2 or G Š D4 , the dihedral group of order 8, or G Š Q the quaternion group. (h) If jGj D 9 then G Š Z9 or G Š Z3  Z3 . (i) If jGj D 10 then G Š Z10 Š Z2  Z5 or G Š D5 , the dihedral group with 10 elements. Recall from Section 10.1 that a finite group of prime order must be cyclic. Hence in the theorem the cases jGj D 2; 3; 5; 7 are handled. We next consider the case where G has order p 2 where p is a prime. Definition 10.5.4. If G is a group then its center denoted Z.G/ is the set of elements in G which commute with everything in G. That is Z.G/ D ¹g 2 G W gh D hg for any h 2 Gº: Lemma 10.5.5. For any group G the following hold: (a) The center Z.G/ is a normal subgroup. (b) G D Z.G/ if and only if G is abelian. (c) If G=Z.G/ is cyclic then G is abelian. Proof. (a) and (b) are direct and we leave them to the exercises. Consider the case where G=Z.G/ is cyclic. Then each coset of Z.G/ has the form g m Z.G/ where g 2 G. Let a; b 2 G. Then since a; b are in cosets of the center we have a D g m u and b D g n v with u; v 2 Z.G/. Then ab D .g m u/.g n v/ D .g m g n /.uv/ D .g n g m /.vu/ D .g n v/.g m u/ D ba since u; v commute with everything. Therefore G is abelian. A p-group is any finite group of prime power order. We need the following. The proof of this is based on what is called the class equation which we will prove in Chapter 13. Lemma 10.5.6. A finite p-group has a nontrivial center of order at least p. Lemma 10.5.7. If jGj D p2 with p a prime then G is abelian and hence G Š Zp2 or G Š Zp  Zp . Proof. Suppose that jGj D p 2 . Then from the previous lemma G has a nontrivial center and hence jZ.G/j D p or jZ.G/j D p 2 . If jZ.G/j D p 2 then G D Z.G/ and G is abelian. If jZ.G/j D p then jG=Z.G/j D p. Since p is a prime this implies that G=Z.G/ is cyclic and hence from Lemma 10.5.5 G is abelian.

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Chapter 10 Normal Subgroups, Factor Groups and Direct Products

Lemma 10.5.7 handles the cases n D 4 and n D 9. Therefore if jGj D 4 we must have G Š Z4 or G Š Z2  Z2 and if jGj D 9 we must have G Š Z9 or G Š Z 3  Z3 . This leaves n D 6; 8; 10. We next handle 6 and 10. Lemma 10.5.8. If G is any group where every nontrivial element has order 2 then G is abelian. Proof. Suppose that g 2 D 1 for all g 2 G. This implies that g D g 1 for all g 2 G. Let a; b be arbitrary elements of G. Then .ab/2 D 1 H) abab D 1 H) ab D b 1 a1 D ba: Therefore a; b commute and G is abelian. Lemma 10.5.9. If jGj D 6 then G Š Z6 or G Š D3 . Proof. Since 6 D 2  3 if G was abelian then G Š Z2  Z3 . Notice that if an abelian group has an element of order m and an element of order n with .n; m/ D 1 then it has an element of order mn. Therefore for 6 if G is abelian there is an element of order 6 and hence G Š Z2  Z3 Š Z6 . Now suppose that G is nonabelian. The nontrivial elements of G have orders 2, 3 or 6. If there is an element of order 6 then G is cyclic and hence abelian. If every element has order 2 then G is abelian. Therefore there is an element of order 3 say g 2 G. The cyclic subgroup hgi D ¹1; g; g 2 º then has index 2 in G and is therefore normal. Let h 2 G with h … hgi. Since g; g 2 both generate hgi we must have jhgijjhhij hgi \ hhi D ¹1º. If h also had order 3 then jhg; hij D jhgi\hhij D 9 which is impossible. Therefore h must have order 2. Since hgi is normal we have h1 gh D g t for t D 1; 2. If h1 gh D g then g; h commute and the group G is abelian. Therefore h1 gh D g 2 D g 1 . It follows that g; h generate a subgroup of G satisfying g 3 D h2 D 1;

h1 gh D g 1 :

This defines a subgroup of order 6 isomorphic to D3 and hence must be all of G. Lemma 10.5.10. If jGj D 10 then G Š Z10 or G Š D5 . Proof. The proof is almost identical to that for n D 6. Since 10 D 2  5 if G were abelian G Š Z2  Z5 D Z10 . Now suppose that G is nonabelian. As for n D 6, G must contain a normal cyclic subgroup of order 5 say hgi D ¹1; g; g 2 ; g 3 ; g 4 º. If h … hgi then exactly as for n D 6 it follows that h must have order 2 and h1 gh D g t for t D 1; 2; 3; 4. If

Section 10.5 Some Properties of Finite Groups

159

h1 gh D g then g; h commute and G is abelian. Notice that h1 D h. Suppose that h1 gh D hgh D g 2 . Then .hgh/3 D .g2 /3 D g 6 D g H) g D h2 gh2 D hg2 h D g4 H) g D 1 which is a contradiction. Similarly hgh D g 3 leads to a contradiction. Therefore h1 gh D g4 D g 1 and g; h generate a subgroup of order 10 satisfying g 5 D h2 D 1I

h1 gh D g1 :

Therefore this is all of G and is isomorphic to D5 . This leaves the case n D 8 that is the most difficult. If jGj D 8 and G is abelian then clearly G Š Z8 or G Š Z4  Z2 or G Š Z2  Z2  Z2 . The proof of Theorem 10.5.3 is then completed with the following. Lemma 10.5.11. If G is a nonabelian group of order 8 then G Š D4 or G Š Q. Proof. The nontrivial elements of G have orders 2, 4 or 8. If there is an element of order 8 then G is cyclic and hence abelian while if every element has order 2 then G is abelian. Hence we may assume that G has an element of order 4 say g. Then hgi has index 2 and is a normal subgroup. Suppose first that G has an element h … hgi of order 2. Then h1 gh D g t for some t D 1; 2; 3: If h1 gh D g then as in the cases 6 and 10, hg; hi defines an abelian subgroup of order 8 and hence G is abelian. If h1 gh D g2 then .h1 gh/2 D .g 2 /2 D g4 D 1 H) g D h2 gh2 D h1 g 2 h D g 4 H) g 3 D 1 contradicting the fact that g has order 4. Therefore h1 gh D g 3 D g 1 . It follows that g; h define a subgroup of order 8 isomorphic to D4 . Since jGj D 8 this must be all of G and G Š D4 . Therefore we may now assume that every element h 2 G with h … hgi has order 4. Let h be such an element. Then h2 has order 2 so h2 2 hgi which implies that h2 D g 2 . This further implies that g 2 is central, that is commutes with everything. Identifying g with i , h with j and g 2 with 1 we get that G is isomorphic to Q completing Lemma 10.5.11 and the proof of Theorem 10.5.3. In principle this type of analysis can be used to determine the structure of any finite group although it quickly becomes impractical. A major tool in this classification is the following important result known as the Sylow theorem which now we just state. We will prove this theorem in Chapter 13. If jGj D p m n with p a prime and .n; p/ D 1 then a subgroup of G of order p m is called a p-Sylow subgroup. It is not clear at first that a group will contain p-Sylow subgroups.

160

Chapter 10 Normal Subgroups, Factor Groups and Direct Products

Theorem 10.5.12 (Sylow theorem). Let jGj D p m n with p a prime and .n; p/ D 1. (a) G contains a p-Sylow subgroup. (b) All p-Sylow subgroups of G are conjugate G. (c) Any p-subgroup of G is contained in a p-Sylow subgroup. (d) The number of p-Sylow subgroups of G is of the form 1 C pk and divides n.

10.6

Exercises

1. Prove that if G is cyclic then any factor group of G is also cyclic. 2. Prove that for any group G the center Z.G/ is a normal subgroup and G D Z.G/ if and only if G is abelian. 3. Let U1 and U2 be subgroups of a group G. Let x; y 2 G. Show: (i) If xU1 D yU2 then U1 D U2 . (ii) Give an example to show that xU1 D U2 x does not imply U1 D U2 . 4. Let U; V be subgroups of a group G. Let x; y 2 G. If UxV \ UyV ¤ ; then UxV D UyV . 5. Let N be a cyclic normal subgroup of the group G. Then all subgroups of N are normal subgroups of G. Give an example to show that the statement is not correct if N is not cyclic. 6. Let N1 and N2 be normal subgroups of G. Show: (i) If all elements in N1 and N2 have finite order, then also the elements of N1 N 2 . e (ii) Let e1 ; e2 2 N. If ni i D 1 for all ni 2 Ni (i D 1; 2), then x e1 e2 D 1 for all x 2 N 1 N2 .

7. Find groups N1 ; N2 and G with N1 G N2 G G, but N1 is not a normal subgroup of G. 8. Let G be a group generated by a and b and let bab 1 D ar and an D 1 for suitable r 2 Z; n 2 N. Show: (i) The subgroup A WD hai is a normal subgroup of G. (ii) G=A D hbAi. (iii) G D ¹b j ai W i; j 2 Zº. 9. Prove that any group of order 24 cannot be simple.

Chapter 11

Symmetric and Alternating Groups

11.1

Symmetric Groups and Cycle Decomposition

Groups most often appear as groups of transformations or permutations on a set. In Galois Theory groups will appear as permutation groups on the zeros of a polynomial. In Section 9.3 we introduced permutation groups and the symmetric group Sn . In this chapter we look more carefully at the structure of Sn and for each n introduce a very important normal subgroup An of Sn called the alternating group on n symbols. Recall that if A is a set, a permutation on A is a one-to-one mapping of A onto itself. The set SA of all permutations on A forms a group under composition called the symmetric group on A. If jAj > 2 then SA is nonabelian. Further if A; B have the same cardinality, then SA Š SB . If jAj D n then jSA j D nŠ and in this case we denote SA by Sn , called the symmetric group on n symbols. For example jS3 j D 6. In Example 9.3.5 we showed that the six elements of S3 can be given by:       1 2 3 1 2 3 1 2 3 1D ; aD ; bD 1 2 3 2 3 1 3 1 2       1 2 3 1 2 3 1 2 3 cD ; dD ; eD : 2 1 3 3 2 1 1 3 2 Further we saw that S3 has a presentation given by S3 D ha; cI a3 D c 2 D 1; ac D ca2 i: By this we mean that S3 is generated by a; c, or that S3 has generators a; c and the whole group and its multiplication table can be generated by using the relations a3 D c 2 D 1, ac D ca2 . In general a permutation group is any subgroup of SA for a set A. For the remainder of this chapter we will only consider finite symmetric groups Sn and always consider the set A as A D ¹1; 2; 3; : : : ; nº. Definition 11.1.1. Suppose that f is a permutation of A D ¹1; 2; : : : ; nº, which has the following effect on the elements of A: There exists an element a1 2 A such that f .a1 / D a2 , f .a2 / D a3 , . . . , f .ak1 / D ak , f .ak / D a1 , and f leaves all other elements (if there are any) of A fixed, i.e., f .aj / D aj for aj ¤ ai , i D 1; 2; : : : ; k. Such a permutation f is called a cycle or a k-cycle.

162

Chapter 11 Symmetric and Alternating Groups

We use the following notation for a k-cycle, f , as given above: f D .a1 ; a2 ; : : : ; ak /: The cycle notation is read from left to right, it says f takes a1 into a2 , a2 into a3 , etc., and finally ak , the last symbol, into a1 , the first symbol. Moreover, f leaves all the other elements not appearing in the representation above fixed. Note that one can write the same cycle in many ways using this type of notation; e.g., f D .a2 ; a3 ; : : : ; ak ; a1 /. In fact any cyclic rearrangement of the symbols gives the same cycle. The integer k is the length of the cycle. Note we allow a cycle to have length 1, i.e., f D .a1 /, for instance, this is just the identity map. For this reason, we will usually designate the identity of Sn by .1/ or just 1. (Of course, it also could be written as .ai / where ai 2 A.) If f and g are two cycles, they are called disjoint cycles if the elements moved by one are left fixed by the other, that is, their representations contain different elements of the set A (their representations are disjoint as sets). Lemma 11.1.2. If f and g are disjoint cycles, then they must commute, that is, fg D gf . Proof. Since the cycles f and g are disjoint, each element moved by f is fixed by g and vice versa. First suppose f .ai / ¤ ai . This implies that g.ai / D ai and f 2 .ai / ¤ f .ai /. But since f 2 .ai / ¤ f .ai /, g.f .ai // D f .ai /. Thus .fg/.ai / D f .g.ai // D f .ai / while .gf /.ai / D g.f .ai // D f .ai /. Similarly if g.aj / ¤ aj , then .fg/.aj / D .gf /.aj /. Finally, if f .ak / D ak and g.ak / D ak then clearly .fg/.ak / D ak D .gf /.ak /. Thus gf D fg. Before proceeding further with the theory, let us consider a specific example. Let A D ¹1; 2; : : : ; 8º and let   1 2 3 4 5 6 7 8 f D : 2 4 6 5 1 7 3 8 We pick an arbitrary number from the set A, say 1. Then f .1/ D 2, f .2/ D 4, f .4/ D 5, f .5/ D 1. Now select an element from A not in the set ¹1; 2; 4; 5º, say 3. Then f .3/ D 6, f .6/ D 7, f .7/ D 3. Next select any element of A not occurring in the set ¹1; 2; 4; 5º [ ¹3; 6; 7º. The only element left is 8, and f .8/ D 8. It is clear that we can now write the permutation f as a product of cycles: f D .1; 2; 4; 5/.3; 6; 7/.8/

Section 11.1 Symmetric Groups and Cycle Decomposition

163

where the order of the cycles is immaterial since they are disjoint and therefore commute. It is customary to omit such cycles as .8/ and write f simply as f D .1245/.367/ with the understanding that the elements of A not appearing are left fixed by f . It is not difficult to generalize what was done here for a specific example, and show that any permutation f can be written uniquely, except for order, as a product of disjoint cycles. Thus let f be a permutation on the set A D ¹1; 2; : : : ; nº, and let a1 2 A. Let f .a1 / D a2 , f 2 .a1 / D f .a2 / D a3 , etc., and continue until a repetition is obtained. We claim that this first occurs for a1 , that is, the first repetition is say f k .a1 / D f .ak / D akC1 D a1 . For suppose the first repetition occurs at the k-th iterate of f and f k .a1 / D f .ak / D akC1 ; and akC1 D aj , where j < k. Then f k .a1 / D f j 1 .a1 /; and so f kj C1 .a1 / D a1 . However, k j C1 < k if j ¤ 1, and we assumed that the first repetition occurred for k. Thus, j D 1 and so f does cyclically permute the set ¹a1 ; a2 ; : : : ; ak º. If k < n, then there exists b1 2 A such that b1 … ¹a1 ; a2 ; : : : ; ak º and we may proceed similarly with b1 . We continue in this manner until all the elements of A are accounted for. It is then seen that f can be written in the form f D .a1 ; : : : ; ak /.b1 ; : : : ; b` /.c1 ; : : : ; cm /    .h1 ; : : : ; h t /: Note that all powers f i .a1 / belong to the set ¹a1 D f 0 .a1 / D f k .a1 /; a2 D f 1 .a1 /; : : : ; ak D f k1 .a1 /º, all powers f i .b1 / belong to the set ¹b1 D f 0 .b1 / D f ` .b1 /; b2 D f 1 .b1 /; : : : ; b` D f `1 .b1 /º; : : : . Here, by definition, b1 is the smallest element in ¹1; 2; : : : ; nº which does not belong to ¹a1 D f 0 .a1 / D f k .a1 /; a2 D f 1 .a1 /; : : : ; ak D f k1 .a1 /º, c1 is the smallest element in ¹1; 2; : : : ; nº which does not belong to ¹a1 D f 0 .a1 / D f k .a1 /; a2 D f 1 .a1 /; : : : ; ak D f k1 .a1 /º [ ¹b1 D f 0 .b1 / D f ` .b1 /; b2 D f 1 .b1 /; : : : ; b` D f `1 .b1 /º: Therefore by construction, all the cycles are disjoint. From this it follows that k C ` C m C    C t D n. It is clear that this factorization is unique except for the order of the factors since it tells explicitly what effect f has on each element of A. In summary we have proven the following result. Theorem 11.1.3. Every permutation of Sn can be written uniquely as a product of disjoint cycles (up to order).

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Chapter 11 Symmetric and Alternating Groups

Example 11.1.4. The elements of S3 can be written in cycle notation as 1 D .1/; .1; 2/; .1; 3/; .2; 3/; .1; 2; 3/; .1; 3; 2/. This is the largest symmetric group which consists entirely of cycles. In S4 , for example, the element .1; 2/.3; 4/ is not a cycle. Suppose we multiply two elements of S3 say .1; 2/ and .1; 3/. In forming the product or composition here, we read from right to left. Thus to compute .1; 2/.1; 3/: We note the permutation .1; 3/ takes 1 into 3 and then the permutation .1; 2/ takes 3 into 3 so the composite .1; 2/.1; 3/ takes 1 into 3. Continuing the permutation .1; 3/ takes 3 into 1 and then the permutation .1; 2/ takes 1 into 2, so the composite .1; 2/.1; 3/ takes 3 into 2. Finally .1; 3/ takes 2 into 2 and then .1; 2/ takes 2 into 1 so .1; 2/.1; 3/ takes 2 into 1. Thus we see .1; 2/.1; 3/ D .1; 3; 2/: As another example of this cycle multiplication consider the product in S5 , .1; 2/.2; 4; 5/.1; 3/.1; 2; 5/: Reading from right to left 1 7! 2 7! 2 7! 4 7! 4 so 1 7! 4. Now 4 7! 4 7! 4 7! 5 7! 5 so 4 7! 5. Next 5 7! 1 7! 3 7! 3 7! 3 so 5 7! 3. Then 3 7! 3 7! 1 7! 1 7! 2 so 3 7! 2. Finally 2 7! 5 7! 5 7! 2 7! 1, so 2 7! 1. Since all the elements of A D ¹1; 2; 3; 4; 5º have been accounted for, we have .1; 2/.2; 4; 5/.1; 3/.1; 2; 5/ D .1; 4; 5; 3; 2/: Let f 2 Sn . If f is a cycle of length 2, i.e., f D .a1 ; a2 / where a1 ; a2 2 A, then f is called a transposition. Any cycle can be written as a product of transpositions, namely .a1 ; : : : ; ak / D .a1 ; ak /.a1 ; ak1 /    .a1 ; a2 /: From Theorem 11.1.3 any permutation can be written in terms of cycles, but from the above any cycle can be written as a product of transpositions. Thus we have the following result. Theorem 11.1.5. Let f 2 Sn be any permutation of degree n. Then f can be written as a product of transpositions.

11.2

Parity and the Alternating Groups

If f is a permutation with a cycle decomposition .a1 ; : : : ; ak /.b1 ; : : : ; bj /    .m1 ; : : : ; m t / then f can be written as a product of W .f / D .k  1/ C .j  1/ C    C .t  1/

Section 11.2 Parity and the Alternating Groups

165

transpositions. The number W .f / is uniquely associated with the permutation f since f is uniquely represented (up to order) as a product of disjoint cycles. However, there is nothing unique about the number of transpositions occurring in an arbitrary representation of f as a product of transpositions. For example in S3 .1; 3; 2/ D .1; 2/.1; 3/ D .1; 2/.1; 3/.1; 2/.1; 2/; since .1; 2/.1; 2/ D .1/, the identity permutation of S3 . Although the number of transpositions is not unique in the representation of a permutation, f , as a product of transpositions, we shall show, however, that the parity (even or oddness) of that number is unique. Moreover, this depends solely on the number W .f / uniquely associated with the representation of f . More explicitly, we have the following result. Theorem 11.2.1. If f is a permutation written as a product of disjoint cycles and if W .f / is the associated integer given above, then if W .f / is even (odd) any representation of f as a product of transpositions must contain an even (odd) number of transpositions. Proof. We first observe the following: .a; b/.b; c1 ; : : : ; c t /.a; b1 ; : : : ; bk / D .a; b1 ; : : : ; bk ; b; c1 ; : : : ; c t /; .a; b/.a; b1 ; : : : ; bk ; b; c1 ; : : : ; c t / D .a; b1 ; : : : ; bk /.b; c1 ; : : : ; c t /: Suppose now that f is represented as a product of disjoint cycles, where we include all the 1-cycles of elements of A which f fixes, if any. If a and b occur in the same cycle in this representation for f , f D    .a; b1 ; : : : ; bk ; b; c1 ; : : : ; c t /    ; then in the computation of W .f / this cycle contributes k C t C 1. Now consider .a; b/f . Since the cycles are disjoint and disjoint cycles commute, .a; b/f D    .a; b/.a; b1 ; : : : ; bk ; b; c1 ; : : : ; c t /    since neither a nor b can occur in any factor of f other than .a; b1 ; : : : ; bk ; b; c1 ; : : : ; c t /. So that .a; b/ cancels out and we find that .a; b/f D    .b; c1 ; : : : ; c t /.a; b1 ; : : : ; bk /    . Since W ..b; c1 ; : : : ; c t /.a; b1 ; : : : ; bk // D k C t but W .a; b1 ; : : : ; bk ; b; c1 ; : : : ; c t / D k C t C 1, we have W ..a; b/f / D W .f /  1. A similar analysis shows that in the case where a and b occur in different cycles in the representation of f , then W ..a; b/f / D W .f / C 1. Combining both cases, we have W ..a; b/f / D W .f / ˙ 1:

166

Chapter 11 Symmetric and Alternating Groups

Now let f be written as a product of m transpositions, say f D .a1 ; b1 /.a2 ; b2 /    .am ; bm /: Then .am ; bm /    .a2 ; b2 /.a1 ; b1 /f D 1: Iterating this, together with the fact that W .1/ D 0, shows that W .f /.˙1/.˙1/.˙1/    .˙1/ D 0; where there are m terms of the form ˙1. Thus W .f / D .˙1/.˙1/    .˙1/; m times. Note if exactly p are C and q D m  p are  then m D p C q and W .f / D p  q. Hence m  W .f / .mod 2/. Thus, W .f / is even if and only if m is even and this completes the proof. It now makes sense to state the following definition since we know that the parity is indeed unique. Definition 11.2.2. A permutation f 2 Sn is said to be even if it can be written as a product of an even number of transpositions. Similarly, f is called odd if it can be written as a product of an odd number of transpositions. Definition 11.2.3. On the group Sn we define the sign function sgn W Sn ! .Z2 ; C/ by sgn./ D 0 if  is an even permutation and sgn./ D 1 if  is an odd permutation. We note that if f and g are even permutations then so are fg and f 1 and also the identity permutation is even. Further if f is even and g is odd it is clear that fg is odd. From this it is straightforward to establish the following. Lemma 11.2.4. sgn is a homomorphism from Sn onto .Z2 ; C/. We now let An D ¹ 2 Sn W sgn./ D 0º: That is, An is precisely the set of even permutations in Sn . Theorem 11.2.5. For each n 2 N the set An forms a normal subgroup of index 2 in Sn called the alternating group on n symbols. Further jAn j D nŠ 2. Proof. By Lemma 11.2.4 sgn W Sn ! .Z2 ; C/ is a homomorphism. Then ker.sgn/ D An and therefore An is a normal subgroup of Sn . Since im.sgn/ D Z2 we have jim.sgn/j D 2 and hence jSn =An j D 2. Therefore ŒSn W An  D 2. Since jSn j D nŠ then jAn j D nŠ 2 follows from Lagrange’s theorem.

167

Section 11.3 Conjugation in Sn

11.3

Conjugation in Sn

Recall that in a group G two elements x; y 2 G are conjugates if there exists a g 2 G with g 1 xg D y. Conjugacy is an equivalence relation on G. In the symmetric groups Sn it is easy to determine if two elements are conjugates. We say that two permutations in Sn have the same cycle structure if they have the same number of cycles and the lengths are the same. Hence for example in S8 the permutations 1 D .1; 3; 6; 7/.2; 5/ and

2 D .2; 3; 5; 6/.1; 8/

have the same cycle structure. In particular if 1 ; 2 are two permutations in Sn then 1 ; 2 are conjugates if and only if they have the same cycle structure. Therefore in S8 the permutations 1 D .1; 3; 6; 7/.2; 5/ and

2 D .2; 3; 5; 6/.1; 8/

are conjugates. Lemma 11.3.1. Let  D .a11 ; a12 ; : : : ; a1k1 /    .as1 ; as2 ; : : : ; asks /  the image of a be the cycle decomposition of  2 Sn . Let  2 Sn and denote by aij ij under  . Then         1 D .a11 ; a12 ; : : : ; a1k /    .as1 ; as2 ; : : : ; ask /: 1 s

Proof. (a) Consider a11 then operating on the left like functions we have   / D  .a11 / D  .a12 / D a12 :   1 .a11

The same computation then follows for all the symbols aij proving the lemma. Theorem 11.3.2. Two permutations 1 ; 2 2 Sn are conjugates if and only if they are of the same cycle structure. Proof. Suppose that 2 D  1  1 . Then from Lemma 11.3.1 we have that 1 and 2 are of the same cycle structure. Conversely suppose that 1 and 2 are of the same cycle structure. Let 1 D .a11 ; a12 ; : : : ; a1k1 /    .as1 ; as2 ; : : : ; asks / 2 D .b11 ; b12 ; : : : ; b1k1 /    .bs1 ; bs2 ; : : : ; bsks / where we place the cycles of the same length under each other. Let  be the permutation in Sn that maps each symbol in 1 to the digit below it in 2 . Then from Lemma 11.3.1 we have  1  1 D 2 and hence 1 and 2 are conjugate.

168

11.4

Chapter 11 Symmetric and Alternating Groups

The Simplicity of An

A simple group is a group G with no nontrivial proper normal subgroups. Up to this point the only examples we have of simple groups are cyclic groups of prime order. In this section we prove that if n  5 each alternating group An is a simple group. Theorem 11.4.1. For each n  3 each  2 An is a product of cycles of length 3. Proof. Let  2 An . Since  is a product of an even number of transpositions to prove the theorem it suffices to show that if 1 ; 2 are transpositions then 1 2 is a product of 3-cycles. Suppose that a; b; c; d are different digits in ¹1; : : : ; nº. There are three cases to consider. First: Case (1): .a; b/.a; b/ D 1 D .1; 2; 3/0 and hence it is true here. Next: Case (2):

.a; b/.b; c/ D .c; a; b/

and hence it is true here also. Finally: Case (3):

.a; b/.c; d / D .a; b/.b; c/.b; c/.c; d / D .c; a; b/.c; d; b/

since .b; c/.b; c/ D 1. Therefore it is true here also proving the theorem. Now our main result: Theorem 11.4.2. For n  5 the alternating group An is a simple nonabelian group. Proof. Suppose that N is a nontrivial normal subgroup of An with n  5. We show that N D An and hence that An is simple. We claim first that N must contain a 3-cycle. Let 1 ¤  2 N then  is not a transposition since  2 An . Therefore  moves at least 3 digits. If  moves exactly 3 digits then it is a 3-cycle and we are done. Suppose then that  moves at least 4 digits. Let  D 1    r with i disjoint cycles. Case (1): There is a i D .: : : ; a; b; c; d /. Set D .a; b; c/ 2 An . Then   1 D i i1 D .b; c; d /: However from Lemma 11.3.1 .b; c; d / D .ai ; b i ; c i /. Further since  2 N and N is normal we have .  1 1 / D .b; c; d /.a; c; b/ D .a; d; b/: Therefore in this case N contains a 3-cycle.

169

Section 11.4 The Simplicity of An

Case (2): There is a i which is a 3-cycle. Then  D .a; b; c/.d; e; : : :/: Now set D .a; b; d / 2 An and then   1 D .b; c; e/ D .a ; b  ; d  / and

1   1 D .a; b; d /.b; c; e/ D .b; c; e; d; a/ 2 N:

Now use Case (1). Therefore in this case N has a 3-cycle. In the final case  is a disjoint product of transpositions. Case (3):  D .a; b/.c; d /    . Since n  5 there is an e ¤ a; b; c; d . Let

D .a; c; e/ 2 An . Then   1 D .b; d; e1 /

with e1 D e  ¤ b; d:

However .a ; c  ; e  / D .b; d; e1 /. Let D . 1  / 1 . This is in N since N is normal. If e D e1 then D .e; c; a/.b; d; e/ D .a; e; b; d; c/ and we can use Case (1) to get that N contains a 3-cycle. If e ¤ e1 then  D .e; c; a/.b; d; e1 / 2 N and then we can use Case (2) to obtain that N contains a 3-cycle. These three cases show that N must contain a 3-cycle. If N is normal in An then from the argument above N contains a 3-cycle  . However from Theorem 11.3.2 any two 3-cycles in Sn are conjugate. Hence  is conjugate to any other 3-cycle in Sn . Since N is normal and  2 N each of these conjugates must also be in N . Therefore N contains all 3-cycles in Sn . From Theorem 11.4.1 each element of An is a product of 3-cycles. It follows then that each element of An is in N . However since N  An this is only possible if N D An completing the proof. Theorem 11.4.3. Let p be a prime and U  Sp a subgroup. Let  be a transposition and ˛ a p-cycle with ˛;  2 U . Then U D Sp . Proof. Suppose without loss of generality that  D .1; 2/. Since ˛; ˛ 2 ; : : : ; ˛ p1 are p-cycles with no fixed points then there is an i with ˛ i .1/ D 2. Without loss of generality we may then assume that ˛ D .1; 2; a3 ; : : : ; ap /. Let   1 2 a3    ap : D 1 2 3  p Then from Lemma 10.3.4 we have  ˛ 1 D .1; 2; : : : ; p/: Further .1; 2/ 1 D .1; 2/. Hence U1 D U 1 contains .1; 2/ and .1; 2; : : : ; p/.

170

Chapter 11 Symmetric and Alternating Groups

Now we have .1; 2; : : : ; p/.1; 2/.1; 2; : : : ; p/1 D .2; 3/ 2 U1 : Analogously

.1; 2; : : : ; p/.2; 3/.1; 2; : : : ; p/1 D .3; 4/ 2 U1 ;

and so on until .1; 2; : : : ; p/.p  2; p  1/.1; 2; : : : ; p/1 D .p  1; p/ 2 U1 : Hence the transpositions .1; 2/; .2; 3/; : : : ; .p  1; p/ 2 U1 . Moreover .1; 2/.2; 3/.1; 2/ D .1; 3/ 2 U1 : In an identical fashion each .1; k/ 2 U1 . Then for any digits s; t we have .1; s/.1; t /.1; s/ D .s; t / 2 U1 : Therefore U1 contains all the transpositions of Sp and hence U1 D Sp . Since U D U1  1 we must have U D Sp also.

11.5

Exercises

1. Show that for n  3 the group An is generated by ¹.12k/ W k  3º. 2. Let D .k1 ; : : : ; ks / 2 Sn be a permutation. Show that theorder of is the least common multiple of k1 ; : : : ; ks . Compute the order of  D 12 26 35 41 53 64 77 2 S7 . 3. Let G D S4 . (i) Determine a noncyclic subgroup H of order 4 of G. (ii) Show that H is normal. (iii) Show that f .g/.h/ WD ghg1 defines an epimorphism f W G ! Aut.H / for g 2 G and h 2 H . Determine its kernel. 4. Show that all subgroups of order 6 of S4 are conjugate.

Chapter 12

Solvable Groups

12.1

Solvability and Solvable Groups

The original motivation for Galois theory grew out of a famous problem in the theory of equations. This problem was to determine the solvability or insolvability of a polynomial equation of degree 5 or higher in terms of a formula involving the coefficients of the polynomial and only using algebraic operations and radicals. This question arose out of the well-known quadratic formula. The ability to solve quadratic equations and in essence the quadratic formula was known to the Babylonians some 3600 years ago. With the discovery of imaginary numbers, the quadratic formula then says that any degree two polynomial over C can be solved by radicals in terms of the coefficients. In the sixteenth century the Italian mathematician Niccolo Tartaglia discovered a similar formula in terms of radicals to solve cubic equations. This cubic formula is now known erroneously as Cardano’s formula in honor of Cardano, who first published it in 1545. An earlier special version of this formula was discovered by Scipione del Ferro. Cardano’s student Ferrari extended the formula to solutions by radicals for fourth degree polynomials. The combination of these formulas says that polynomial equations of degree four or less over the complex numbers can be solved by radicals. From Cardano’s work until the very early nineteenth century, attempts were made to find similar formulas for degree five polynomials. In 1805 Ruffini proved that fifth degree polynomial equations are insolvable by radicals in general. Therefore there exists no comparable formula for degree 5. Abel in 1825–1826 and Galois in 1831 extended Ruffini’s result and proved the insolubility by radicals for all degrees five or greater. In doing this, Galois developed a general theory of field extensions and its relationship to group theory. This has come to be known as Galois theory and is really the main focus of this book. The solution of the insolvability of the quintic and higher involved a translation of the problem into a group theory setting. For a polynomial equation to be solvable by radicals its corresponding Galois group (a concept we will introduce in Chapter 16) must be a solvable group. This is a group with a certain defined structure. In this chapter we introduce and discuss this class of groups.

172

12.2

Chapter 12 Solvable Groups

Solvable Groups

A normal series for a group G is a finite chain of subgroups beginning with G and ending with the identity subgroup ¹1º G D G0 G1 G2    ¹1º in which each Gi C1 is a proper normal subgroup of Gi . The factor groups Gi =Gi C1 are called the factors of the series and n is the length of the series. Definition 12.2.1. A group G is solvable if it has a normal series with abelian factors, that is Gi =Gi C1 is abelian for all i D 0; 1; : : : ; n. Such a normal series is called a solvable series. If G is an abelian group then G D G0 ¹1º provides a solvable series. Hence any abelian group is solvable. Further the symmetric group S3 on 3-symbols is also solvable however nonabelian. Consider the series S3 A3  ¹1º: Since jS3 j D 6 we have jA3 j D 3 and hence A3 is cyclic and therefore abelian. Further jS3 =A3 j D 2 and hence the factor group S3 =A3 is also cyclic and hence abelian. Therefore the series above gives a solvable series for S3 . Lemma 12.2.2. If G is a finite solvable group then G has a normal series with cyclic factors. Proof. If G is a finite solvable group then by definition it has a normal series with abelian factors. Hence to prove the lemma it suffices to show that a finite abelian group has a normal series with cyclic factors. Let A be a nontrivial finite abelian group. We do an induction on the order of A. If jAj D 2 then A itself is cyclic and the result follows. Suppose that jAj > 2. Choose an 1 ¤ a 2 A. Let N D hai so that N is cyclic. Then we have the normal series A N ¹1º with A=N abelian. Further A=N has order less than A so A=N has a normal series with cyclic factors and the result follows. Solvability is preserved under subgroups and factor groups. Theorem 12.2.3. Let G be a solvable group. Then: (1) Any subgroup H of G is also solvable. (2) Any factor group G=N of G is also solvable.

173

Section 12.2 Solvable Groups

Proof. (1) Let G be a solvable group and suppose that G D G0 G1    Gr ¹1º is a solvable series for G. Hence Gi C1 is a normal subgroup of Gi for each i and the factor group Gi =Gi C1 is abelian. Now let H be a subgroup of G and consider the chain of subgroups H D H \ G0 H \ G1    H \ Gr ¹1º: Since GiC1 is normal in Gi we know that H \ Gi C1 is normal in H \ Gi and hence this gives a finite normal series for H . Further from the second isomorphism theorem we have for each i , .H \ Gi /=.H \ Gi C1 / D .H \ Gi /=..H \ Gi / \ Gi C1 / Š .H \ Gi /Gi C1 =Gi C1  Gi =Gi C1 : However Gi =GiC1 is abelian so each factor in the normal series for H is abelian. Therefore the above series is a solvable series for H and hence H is also solvable. (2) Let N be a normal subgroup of G. Then from (1) N is also solvable. As above let G D G0 G1    Gr ¹1º be a solvable series for G. Consider the chain of subgroups G=N D G0 N=N G1 N=N    Gr N=N N=N D ¹1º: Let m 2 Gi1 ; n 2 N . Then since N is normal in G, .mn/1 Gi N.mn/ D n1 m1 Gi mnN D n1 Gi nN D n1 NGi D NGi D Gi N: It follows that Gi C1 N is normal in Gi N for each i and therefore the series for G=N is a normal series. Further, again from the isomorphism theorems .Gi N=N /=.Gi C1 N=N / Š Gi =.Gi \ Gi C1 N / Š .Gi =Gi C1 =..Gi \ Gi C1 N /=Gi C1 /: However the last group .Gi =Gi C1 =..Gi \ Gi C1 N /=Gi C1 / is a factor group of the group Gi =GiC1 which is abelian. Hence this last group is also abelian and therefore each factor in the normal series for G=N is abelian. Hence this series is a solvable series and G=N is solvable.

174

Chapter 12 Solvable Groups

The following is a type of converse of the above theorem. Theorem 12.2.4. Let G be group and H a normal subgroup of G. If both H and G=H are solvable then G is solvable. Proof. Suppose that H D H0 H1    Hr ¹1º G=H D G0 =H G1 =H    Gs =H H=H D ¹1º are solvable series for H and G=H respectively. Then G D G0 G1    Gs D H H1    ¹1º gives a normal series for G. Further from the isomorphism theorems again Gi =Gi C1 Š .Gi =H /=.Gi C1 =H / and hence each factor is abelian. Therefore this is a solvable series for G and hence G is solvable. This theorem allows us to prove that solvability is preserved under direct products. Corollary 12.2.5. Let G and H be solvable groups. Then their direct product G  H is also solvable. Proof. Suppose that G and H are solvable groups and K D G  H . Recall from Chapter 10 that G can be considered as a normal subgroup of K with K=G Š H . Therefore G is a solvable subgroup of K and G=K is a solvable quotient. It follows then from Theorem 12.2.4 that K is solvable. We saw that the symmetric group S3 is solvable. However the following theorem shows that the symmetric group Sn is not solvable for n  5. This result will be crucial to the proof of the insolvability of the quintic and higher. Theorem 12.2.6. For n  5 the symmetric group Sn is not solvable. Proof. For n  5 we saw that the alternating group An is simple. Further An is nonabelian. Hence An cannot have a nontrivial normal series and so no solvable series. Therefore An is not solvable. If Sn were solvable for n  5 then from Theorem 12.2.3 An would also be solvable. Therefore Sn must also be nonsolvable for n  5. In general for a simple, solvable group we have the following. Lemma 12.2.7. If a group G is both simple and solvable then G is cyclic of prime order.

Section 12.3 The Derived Series

175

Proof. Suppose that G is a nontrivial simple, solvable group. Since G is simple the only normal series for G is G D G0 ¹1º. Since G is solvable the factors are abelian and hence G is abelian. Again since G is simple G must be cyclic. If G were infinite then G Š .Z; C/. However then 2Z is a proper normal subgroup, a contradiction. Therefore G must be finite cyclic. If the order were not prime then for each proper divisor of the order there would be a nontrivial proper normal subgroup. Therefore G must be of prime order. In general a finite p-group is solvable. Theorem 12.2.8. A finite p-group G is solvable. Proof. Suppose that jGj D p n . We do this by induction on n. If n D 1 then jGj D p and G is cyclic, hence abelian and therefore solvable. Suppose that n > 1. Then as used previously G has a nontrivial center Z.G/. If Z.G/ D G then G is abelian and hence solvable. If Z.G/ ¤ G then Z.G/ is a finite p-group of order less than pn . From our inductive hypothesis Z.G/ must be solvable. Further G=Z.G/ is then also a finite p-group of order less than p n so it is also solvable. Hence Z.G/ and G=Z.G/ are both solvable so from Theorem 12.2.4 G is solvable.

12.3

The Derived Series

Let G be a group and let a; b 2 G. The product aba1 b 1 is called the commutator of a and b. We write Œa; b D aba1 b 1 . Clearly Œa; b D 1 if and only if a and b commute. Definition 12.3.1. Let G 0 be the subgroup of G which is generated by the set of all commutators G 0 D gp.¹Œx; y W x; y 2 Gº/: G 0 is called the commutator or (derived) subgroup of G. We sometimes write G 0 D ŒG; G. Theorem 12.3.2. For any group G the commutator subgroup G 0 is a normal subgroup of G and G=G 0 is abelian. Further if H is a normal subgroup of G then G=H is abelian if and only if G 0  H . Proof. The commutator subgroup G 0 consists of all finite products of commutators and inverses of commutators. However Œa; b1 D .aba1 b 1 /1 D bab 1 a1 D Œb; a

176

Chapter 12 Solvable Groups

and so the inverse of a commutator is once again a commutator. It then follows that G 0 is precisely the set of all finite products of commutators, i.e., G 0 is the set of all elements of the form h 1 h2    h n where each hi is a commutator of elements of G. If h D Œa; b for a; b 2 G, and x 2 G, xhx 1 D Œxax 1 ; xbx 1  is again a commutator of elements of G. Now from our previous comments, an arbitrary element of G 0 has the form h1 h2    hn , where each hi is a commutator. Thus x.h1 h2    hn /x 1 D .xh1 x 1 /.xh2 x 1 /    .xhn x 1 / and, since by the above each xhi x 1 is a commutator, x.h1 h2    hn /x 1 2 G 0 . It follows that G 0 is a normal subgroup of G. Consider the factor group G=G 0 . Let aG 0 and bG 0 be any two elements of G=G 0 . Then ŒaG 0 ; bG 0  D aG 0  bG 0  .aG 0 /1  .bG 0 /1 D aG 0  bG 0  a1 G 0  b 1 G 0 D aba1 b 1 G 0 D G 0 since Œa; b 2 G 0 . In other words, any two elements of G=G 0 commute and therefore G=G 0 is abelian. Now let N be a normal subgroup of G with G=N abelian. Let a; b 2 G then aN and bN commute since G=N is abelian. Therefore ŒaN; bN  D aN bNa1 N b 1 N D aba1 b 1 N D N: It follows that Œa; b 2 N . Therefore all commutators of elements in G lie in N and therefore the commutator subgroup G 0  N . From the second part of Theorem 12.3.2 we see that G 0 is the minimal normal subgroup of G such that G=N is abelian. We call G=G 0 D Gab the abelianization of G. We consider next the following inductively defined sequence of subgroups of an arbitrary group G called the derived series. Definition 12.3.3. For an arbitrary group G define G .0/ D G and G .1/ D G 0 and then inductively G .nC1/ D .G .n/ /0 . That is G .nC1/ is the commutator subgroup or derived group of G .n/ . The chain of subgroups G D G .0/ G .1/    G .n/    is called the derived series for G. Notice that since G .i C1/ is the commutator subgroup of G .i/ we have G .i/ =G .i C1/ is abelian. If the derived series was finite then G would have a normal series with abelian factors and hence be solvable. The converse is also true and characterizes solvable groups in terms of the derived series.

Section 12.4 Composition Series and the Jordan–Hölder Theorem

177

Theorem 12.3.4. A group G is solvable if and only if its derived series is finite. That is there exists an n such that G .n/ D ¹1º. Proof. If G .n/ D ¹1º for some n then as explained above the derived series provides a solvable series for G and hence G is solvable. Conversely suppose that G is solvable and let G D G0 G1    Gr D ¹1º be a solvable series for G. We claim first that Gi G .i/ for all i . We do this by induction on r. If r D 0 then G D G0 D G .0/ . Suppose that Gi G .i/ . Then Gi0 .G .i/ /0 D G .i C1/ . Since Gi =Gi C1 is abelian it follows from Theorem 12.3.2 that GiC1 Gi0 . Therefore Gi C1 G .i C1/ establishing the claim. Now if G is solvable from the claim we have that Gr G .r/ . However Gr D ¹1º and therefore G .r/ D ¹1º proving the theorem. The length of the derived series is called the solvability length of a solvable group G. The class of solvable groups of class c consists of those solvable groups of solvability length c or less.

12.4

Composition Series and the Jordan–Hölder Theorem

The concept of a normal series is extremely important in the structure theory of groups. This is especially true for finite groups. If G D G0 G1    ¹1º G D H0 H1    ¹1º are two normal series for the group G then the second is a refinement of the first if all the terms of the second occur in the first series. Further, two normal series are called equivalent or (isomorphic) if there exists a 1-1 correspondence between the factors (hence the length must be the same) of the two series such that the corresponding factors are isomorphic. Theorem 12.4.1 (Schreier’s theorem). Any two normal series for a group G have equivalent refinements. Proof. Consider two normal series for G G D G0 G1    Gs ¹1º D GsC1 G D H0 H1    H t ¹1º D G tC1 :

178

Chapter 12 Solvable Groups

Now define Gij D .Gi \ Hj /Gi C1 ;

j D 0; 1; 2; : : : ; t C 1;

Hj i D .Gi \ Hj /Hj C1 ;

i D 0; 1; 2; : : : ; s C 1:

Then we have G D G00 G01    G0;sC1 D G1 D G10    G1;sC1 D G2    G t;sC1 D ¹eº; and G D H00 H01    H0;tC1 D H1 D H10    H1;tC1 D H2    Hs;tC1 D ¹eº: Now applying the third isomorphism theorem to the groups Gi , Hj , Gi C1 , Hj C1 , we have that Gi;j C1 D .Gi \ Hj C1 /Gi C1 is a normal subgroup of Gi;j D .Gi \ Hj /GiC1 and Hj;i C1 D .Gi C1 \ Hj /Hj C1 is a normal subgroup of Hj;i D .Gi \ Hj /Hj C1 . Furthermore, also Gij =Gi;j C1 Š Hj i =Hj;i C1 : Thus the above two are normal series which are refinements of the two given series and they are equivalent. A proper normal subgroup N of a group G is called maximal in G if there does not exist any normal subgroup N  M  G with all inclusions proper. This is the group theoretic analog of a maximal ideal. An alternative characterization is the following. N is a maximal normal subgroup of G if and only if G=N is simple. A normal series where each factor is simple can have no refinements. Definition 12.4.2. A composition series for a group G is a normal series where all the inclusions are proper and such that Gi C1 is maximal in Gi . Equivalently a normal series where each factor is simple. It is possible that an arbitrary group does not have a composition series or even if it does have one, a subgroup of it may not have one. Of course, a finite group does have a composition series. In the case in which a group, G, does have a composition series the following important theorem, called the Jordan–Hölder theorem, provides a type of unique factorization. Theorem 12.4.3 (Jordan–Hölder theorem). If a group G has a composition series, then any two composition series are equivalent, that is the composition factors are unique.

Section 12.5 Exercises

179

Proof. Suppose we are given two composition series. Applying Theorem 12.4.1 we get that the two composition series have equivalent refinements. But the only refinement of a composition series is one obtained by introducing repetitions. If in the 1-1 correspondence between the factors of these refinements, the paired factors equal to ¹eº are disregarded, that is if we drop the repetitions, we get clearly that the original composition series are equivalent. We remarked in Chapter 10 that the simple groups are important because they play a role in finite group theory somewhat analogous to that of the primes in number theory. In particular, an arbitrary finite group, G, can be broken down into simple components. These uniquely determined simple components are, according to the Jordan–Hölder theorem, the factors of a composition series for G.

12.5

Exercises

1. Let K be a field and ´ a x y! μ 0 b z W a; b; c; x; y; z 2 K; abc ¤ 0 : GD 0 0 c Show that G is solvable. 2. A group G is called polycyclic if it has a normal series with cyclic factors. Show: (i) Each subgroup and each factor group of a polycyclic group is polycyclic. (ii) In a polycyclic group each normal series has the same number of infinite cyclic factors. 3. Let G be a group. Show: (i) If G is finite and solvable, then G is polycyclic. (ii) If G is polycyclic, then G is finitely generated. (iii) The group .Q; C/ is solvable, but not polycyclic. 4. Let N1 and N2 be normal subgroups of G. Show: (i) If N1 and N2 are solvable, then also N1 N2 is a solvable normal subgroup of G. (ii) Is (i) still true, if we replace “solvable” by “abelian”? 5. Let N1 ; : : : ; N t be normal subgroups of a group G. If all factor groups G=Ni are solvable, then also G=.N1 \    \ N t / is solvable.

Chapter 13

Groups Actions and the Sylow Theorems

13.1

Group Actions

A group action of a group G on a set A is a homomorphism from G into SA the symmetric group on A. We say that G acts on A. Hence G acts on A if to each g 2 G corresponds a permutation g W A ! A such that (1) g1 .g2 .a// D g1 g2 .a/ for all g1 ; g2 2 G and for all a 2 A (2) 1.a/ D a for all a 2 A. For the remainder of this chapter if g 2 G and a 2 A we will write ga for g .a/. Group actions are an extremely important idea and we use this idea in the present chapter to prove several fundamental results in group theory. If G acts on the set A then we say that two elements a1 ; a2 2 A are congruent under G if there exists a g 2 G with ga1 D a2 . The set Ga D ¹a1 2 A W a1 D ga for some g 2 Gº is called the orbit of a. It consists of elements congruent to a under G. Lemma 13.1.1. If G acts on A then congruence under G is an equivalence relation on A. Proof. Any element a 2 A is congruent to itself via the identity map and hence the relation is reflexive. If a1  a2 so that ga1 D a2 for some g 2 G then g 1 a2 D a1 and so a2  a1 and the relation is symmetric. Finally is g1 a1 D a2 and g2 a2 D a3 then g2 g1 a1 D a3 and relation is transitive. Recall that the equivalence classes under an equivalence relation partition a set. For a given a 2 A its equivalence class under this relation is precisely its orbit as defined above. Corollary 13.1.2. If G acts on the set A then the orbits under G partition the set A. We say that G acts transitively on A if any two elements of A are congruent under G. That is the action is transitive if for any a1 ; a2 2 A there is some g 2 G such that ga1 D a2 .

Section 13.2 Conjugacy Classes and the Class Equation

181

If a 2 A the stabilizer of a consists of those g 2 G that fix a. Hence StabG .a/ D ¹g 2 G W ga D aº: The following lemma is easily proved and left to the exercises. Lemma 13.1.3. If G acts on A then for any a 2 A the stabilizer StabG .a/ is a subgroup of G. We now prove the crucial theorem concerning group actions. Theorem 13.1.4. Suppose that G acts on A and a 2 A. Let Ga be the orbit of a under G and StabG .a/ its stabilizer. Then jG W StabG .a/j D jGa j: That is the size of the orbit of a is the index of its stabilizer in G. Proof. Suppose that g1 ; g2 2 G with g1 StabG .a/ D g2 StabG .a/, that is they define the same left coset of the stabilizer. Then g21 g1 2 StabG .a/. This implies that g21 g1 a D a so that g2 a D g1 a. Hence any two elements in the same left coset of the stabilizer produce the same image of a in Ga . Conversely if g1 a D g2 then g1 ; g2 define the same left coset of StabG .a/. This shows that there is a one-to-one correspondence between left cosets of StabG .a/ and elements of Ga . It follows that the size of Ga is precisely the index of the stabilizer. We will use this theorem repeatedly with different group actions to obtain important group theoretic results.

13.2

Conjugacy Classes and the Class Equation

In Section 10.5 we introduced the center of a group Z.G/ D ¹g 2 G W gg1 D g1 g for all g1 2 Gº; and showed that it is a normal subgroup of G. We then used this normal subgroup in conjunction with what we called the class equation to show that any finite p-group has a nontrivial center. In this section we use group actions to derive the class equation and prove the result for finite p-groups. Recall that if G is a group then two elements g1 ; g2 2 G are conjugate if there exists a g 2 G with g 1 g1 g D g2 . We saw that conjugacy is an equivalence relation on G. For g 2 G its equivalence class is called its conjugacy class that we will denote by Cl.g/. Thus Cl.g/ D ¹g1 2 G W g1 is conjugate to gº:

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Chapter 13 Groups Actions and the Sylow Theorems

If g 2 G then its centralizer CG .g/ is the set of elements in G that commute with g: CG .g/ D ¹g1 2 G W gg1 D g1 gº: Theorem 13.2.1. Let G be a finite group and g 2 G. Then the centralizer of g is a subgroup of G and jG W CG .g/j D jCl.g/j: That is the index of the centralizer of g is the size of its conjugacy class. In particular for a finite group the size of each conjugacy class divides the order of the group. Proof. Let the group G act on itself by conjugation. That is g.g1 / D g 1 g1 g. It is easy to show that this is an action on the set G (see exercises). For g 2 G its orbit under this action is precisely its conjugacy class Cl.g/ and the stabilizer is its centralizer CG .g/. The statements in the theorem then follow directly from Theorem 13.1.4. For any group G, since conjugacy is an equivalence relation, the conjugacy classes partition G. Hence [ P GD Cl.g/ g2G

where this union is a disjoint union. It follows that X jCl.g/j jGj D g2G

where this sum is taken over distinct conjugacy classes. If Cl.g/ D ¹gº that is the conjugacy class of g is g alone then CG .g/ D G so that g commutes with all of G. Therefore in this case g 2 Z.G/. This is true for every element of the center and therefore [ P Cl.g/ G D Z.G/ [ g…Z.G/

where again the second union is a disjoint union. The size of G is then the sum of these disjoint pieces so X jCl.g/j: jGj D jZ.G/j C g…Z.G/

However from Theorem 13.2.1 jCl.g/j D jG W CG .g/j so the equation above becomes X jG W CG .g/j: jGj D jZ.G/j C g…Z.G/

This is known as the class equation.

183

Section 13.3 The Sylow Theorems

Theorem 13.2.2 (class equation). Let G be a finite group. Then X jG W CG .g/j jGj D jZ.G/j C g…Z.G/

where the sum is taken over the distinct centralizers. As a first application we prove the result that finite p-groups have nontrivial centers. Theorem 13.2.3. Let G be a finite p-group. Then G has a nontrivial center. Proof. Let G be a finite p-group so that jGj D p n for some n and consider the class equation X jG W CG .g/j: jGj D jZ.G/j C g…Z.G/

Since jG W CG .g/j divides jGj for each g 2 G we must have that pjjG W CG .g/j for each g 2 G. Further pjjGj. Therefore p must divide jZ.G/j and hence jZ.G/j D p m for some m  1 and therefore Z.G/ is nontrivial. The idea of conjugacy and the centralizer of an element can be extended to subgroups. If H1 ; H2 are subgroups of a group G then H1 ; H2 are conjugate if there exists a g 2 G such that g 1 H1 g D H2 . As for elements conjugacy is an equivalence relation on the set of subgroups of G. If H  G is a subgroup then its conjugacy class consists of all the subgroups of G conjugate to it. The normalizer of H is NG .H / D ¹g 2 G W g1 Hg D H º: As for elements let G act on the set of subgroups of G by conjugation. That is for g 2 G the map is given by H 7! g 1 Hg. For H  G the stabilizer under this action is precisely the normalizer. Hence exactly as for elements we obtain the following theorem. Theorem 13.2.4. Let G be a group and H  G a subgroup. Then the normalizer NG .H / of H is a subgroup of G, H is normal in NG .H / and jG W NG .H /j D number of conjugates of H in G:

13.3

The Sylow Theorems

If G is a finite group and H  G is a subgroup then Lagrange’s theorem guarantees that the order of H divides the order of G. However the converse of Lagrange’s theorem is false. That is, if G is a finite group of order n and if d jn, then G need

184

Chapter 13 Groups Actions and the Sylow Theorems

not contain a subgroup of order d . If d is a prime p or a power of a prime p e , however, then we shall see that G must contain subgroups of that order. In particular, we shall see that if p d is the highest power of p that divides n, than all subgroups of that order are actually conjugate, and we shall finally get a formula concerning the number of such subgroups. These theorems constitute the Sylow theorems which we will examine in this section. First we give an example where the converse of Lagrange’s theorem is false. Lemma 13.3.1. The alternating group on 4 symbols A4 has order 12 but has no subgroup of order 6. Proof. Suppose that there exists a subgroup U  A4 with jU j D 6. Then jA4 W U j D 2 since jA4 j D 12 and hence U is normal in A4 . Now id, .1; 2/.3; 4/, .1; 3/.2; 4/, .1; 4/.2; 3/ are in A4 . These each have order 2 and commute so they form a subgroup V  A4 of order 4. This subgroup V Š Z2  Z2 . Then jV jjU j 46 D : 12 D jA4 j  jV U j D jV \ U j jV \ U j It follows that V \ U ¤ ¹1º and since U is normal we have that V \ U is also normal in A4 . Now .1; 2/.3; 4/ 2 V and by renaming the entries in V if necessary we may assume that it is also in U so that .1; 2/.3; 4/ 2 V \ U . Since .1; 2; 3/ 2 A4 we have .3; 2; 1/.1; 2/.3; 4/.1; 2; 3/ D .1; 3/.2; 4/ 2 V \ U and then .3; 2; 1/.1; 4/.2; 3/.1; 2; 3/ D .1; 2/.3; 4/ 2 V \ U: But then V  V \ U and so V  U . But this is impossible since jV j D 4 which doesn’t divide jU j D 6. Definition 13.3.2. Let G be a finite group with jGj D n and let p be a prime such that p a jn but no higher power of p divides n. A subgroup of G of order pa is called a p-Sylow subgroup. It is not a clear that a p-Sylow subgroup must exist. We will prove that for each pjn a p-Sylow subgroup exists. We first consider and prove a very special case. Theorem 13.3.3. Let G be a finite abelian group and let p be a prime such that pjjGj. Then G contains at least one element of order p.

185

Section 13.3 The Sylow Theorems

Proof. Suppose that G is a finite abelian group of order pn. We use induction on n. If n D 1 then G has order p and hence is cyclic and so has an element of order p. Suppose that the theorem is true for all abelian groups of order pm with m < n and suppose that G has order pn. Suppose that g 2 G. If the order of g is pt for some integer t then g t ¤ 1 and g t has order p proving the theorem in this case. Hence we may suppose that g 2 G has order prime to p and we show that there must be an element whose order is a multiple of p and then use the above argument to get an element of exact order p. Hence we have g 2 G with order m where .m; p/ D 1. Since mjjGj D pn we must have mjn. Since G is abelian hgi is normal and the factor group G=hgi is abelian n of order p. m / < pn. By the inductive hypothesis G=hgi has an element hhgi of order p, h 2 G, and hence hp D g k for some k. g k has order m1 jm and therefore h has order pm1 . Now as above hm1 has order p proving the theorem. Therefore if G is an abelian group and if pjn, then G contains a subgroup of order p, the cyclic subgroup of order p generated by an element a 2 G of order p whose existence is guaranteed by the above theorem. We now present the first Sylow theorem. Theorem 13.3.4 (first Sylow theorem). Let G be a finite group and let pjjGj, then G contains a p-Sylow subgroup, that is a p-Sylow subgroup exists. Proof. Let G be a finite group of order pn and as above we do induction on n. If n D 1 then G is cyclic and G is its own maximal p-subgroup and hence all of G is a p-Sylow subgroup. We assume then that if jGj D pm with m < n then G has a p-Sylow subgroup. Assume that jGj D p t m with .m; p/ D 1. We must show that G contains a subgroup of order p t . If H is a proper subgroup whose index is prime to p then jH j D p t m1 with m1 < m. Therefore by the inductive hypothesis H has a p-Sylow subgroup of order p t . This will also be a subgroup of G and hence a p-Sylow subgroup of G. Therefore we may assume that the index of any proper subgroup H of G must have index divisible by p. Now consider the class equation for G, X jGj D jZ.G/j C jG W CG .g/j: g…Z.G/

By assumption each of the indices are divisible by p and also pjjGj. Therefore pjjZ.G/j. It follows that Z.G/ is a finite abelian group whose order is divisible by p. From Theorem 13.3.3 there exists an element g 2 Z.G/  G of order p. Since g 2 Z.G/ we must have hgi normal in G. The factor group G=hgi then has order p t1 m and by the inductive hypothesis must have a p-Sylow subgroup K of order p t1 and hence of index m. By the correspondence theorem there is a subgroup K of G with hgi  K such that K=H Š K. Therefore jKj D p t and K is a p-Sylow subgroup of G.

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Chapter 13 Groups Actions and the Sylow Theorems

On the basis of this theorem, we can now strengthen the result obtained in Theorem 13.3.3. Theorem 13.3.5 (Cauchy). If G is a finite group and if p is a prime such that pjjGj, then G contains at least one element of order p. Proof. Let P be a p-Sylow subgroup of G, and let jP j D pt . If g 2 P , g ¤ 1, then t 1 the order of g is p t1 . Then g p 1 has order p. We have seen that p-Sylow subgroups exist. We now wish to show that any two p-Sylow subgroups are conjugate. This is the content of the second Sylow theorem. Theorem 13.3.6 (second Sylow theorem). Let G be a finite group and p a prime such that pjjGj. Then any p-subgroup H of G is contained in a p-Sylow subgroup. Further all p-Sylow subgroups of G are conjugate. That is, if P1 and P2 are any two p-Sylow subgroups of G then there exists an a 2 G such that P1 D aP2 a1 . Proof. Let  be the set of p-Sylow subgroups of G and let G act on  by conjugation. This action will of course partition  into disjoint orbits. Let P be a fixed p-Sylow subgroup and P be its orbit under the conjugation action. The size of the orbit is the index of its stabilizer that is jP j D jG W StabG .P /j. Now P  StabG .P / and P is a maximal p-subgroup of G. It follows that the index of StabG .P / must be prime to p and so the number of p-Sylow subgroups conjugate to P is prime to p. Now let H be a p-subgroup of G and let H act on P by conjugation. P will itself decompose into disjoint orbits under this actions. Further the size of each orbit is an index of a subgroup of H and hence must be a power of p. On the other hand the size of the whole orbit is prime to p. Therefore there must be one orbit that has size exactly 1. This orbit contains a p-Sylow subgroup P 0 and P 0 is fixed by H under conjugation, that is H normalizes P 0 . It follows that HP 0 is a subgroup of G and P 0 is normal in HP 0 . From the second isomorphism theorem we then obtain HP 0 =P 0 Š H=.H \ P 0 /: Since H is a p-group the size of H=.H \ P 0 / is a power of p and therefore so is the size of HP 0 =P 0 . But P 0 is also a p-group so it follows that HP 0 also has order a power of p. Now P 0  HP 0 but P 0 is a maximal p-subgroup of G. Hence HP 0 D P 0 . This is possible only if H  P 0 proving the first assertion in the theorem. Therefore any p-subgroup of G is obtained in a p-Sylow subgroup. Now let H be a p-Sylow subgroup P1 and let P1 act on P . Exactly as in the argument above P1  P 0 where P 0 is a conjugate of P . Since P1 and P 0 are both p-Sylow subgroups they have the same size and hence P1 D P 0 . This implies that P1 is a conjugate of P . Since P1 and P are arbitrary p-Sylow subgroups it follows that all p-Sylow subgroups are conjugate.

Section 13.4 Some Applications of the Sylow Theorems

187

We come now to the last of the three Sylow theorems. This one gives us information concerning the number of p-Sylow subgroups. Theorem 13.3.7 (third Sylow theorem). Let G be a finite group and p a prime such that pjjGj. Then the number of p-Sylow subgroups of G is of the form 1 C pk and divides the order of jGj. It follows that if jGj D p a m with .p; m/ D 1 then the number of p-Sylow subgroups divides m. Proof. Let P be a p-Sylow subgroup and let P act on , the set of all p-Sylow subgroups, by conjugation. Now P normalizes itself so there is one orbit, namely P that has size exactly 1. Every other orbit has size a power of p since the size is the index of a nontrivial subgroup of P and therefore must be divisible by p. Hence the size of the  is 1 C pk.

13.4

Some Applications of the Sylow Theorems

We now give some applications of the Sylow theorems. First we show that the converse of Lagrange’s theorem is true for both general p-groups and for finite abelian groups. Theorem 13.4.1. Let G be a group of order p n . Then G contains at least one normal subgroup of order p m , for each m such that 0  m  n. Proof. We use induction on n. For n D 1 the theorem is trivial. By Lemma 10.5.7 any group of order p 2 is abelian. This together with Theorem 13.3.3 establishes the claim for n D 2. We now assume the theorem is true for all groups G of order p k where 1  k < n, where n > 2. Let G be a group of order p n . From Lemma 10.3.4 G has a nontrivial center of order at least p and hence an element g 2 Z.G/ of order p. Let N D hgi. Since g 2 Z.G/ it follows that N is normal subgroup of order p. Then G=N is of order p n1 , and therefore, contains (by the induction hypothesis) normal subgroups of orders p m1 , for 0  m  1  n  1. These groups are of the form H=N , where the normal subgroup H  G contains N and is of order p m , 1  m  n, because jH j D jN jŒH W N  D jN j  jH=N j. On the basis of the first Sylow theorem we see that if G is a finite group and if p k jjGj, then G must contain a subgroup of order p k . One can actually show that, as in the case of Sylow p-groups, the number of such subgroups is of the form 1 C pt , but we shall not prove this here. Theorem 13.4.2. Let G be a finite abelian group of order n. Suppose that d jn. Then G contains a subgroup of order d .

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Chapter 13 Groups Actions and the Sylow Theorems

Proof. Suppose that n D p1e1    pkek is the prime factorization of n. Then d D f

f

p1 1    pk k for some nonnegative f1 ; : : : ; fk . Now G has p1 -Sylow subgroup H1 of

f order p1e1 . Hence from Theorem 13.4.1 H1 has a subgroup K1 of order p1 1 . Similarly f f there are subgroups K2 ; : : : ; Kk of G of respective orders p2 2 ; : : : ; pk k . Further since the orders are disjoint Ki \ Kj D ¹1º if i ¤ j . It follows that hK1 ; K2 ; : : : ; Kk i has f f order jK1 jjK2 j    jKk j D p1 1    pk k D d .

In Section 10.5 we examined the classification of finite groups of small orders. Here we use the Sylow theorems to extend some of this material further. Theorem 13.4.3. Let p; q be distinct primes with p < q and q not congruent to 1 mod p. Then any group of order pq is cyclic. For example any group of order 15 must be cyclic. Proof. Suppose that jGj D pq with p < q and q not congruent to 1 mod p. The number of q-Sylow subgroups is of the form 1 C qk and divides p. Since q is greater than p this implies that there can be only one and hence there is a normal q-Sylow subgroup H . Since q is a prime, H is cyclic of order q and therefore there is an element g of order q. The number of p-Sylow subgroups is of the form 1 C pk and divides q. Since q is not congruent to 1 mod p this implies that there also can be only one p-Sylow subgroup and hence there is a normal p-Sylow subgroup K. Since p is a prime K is cyclic of order p and therefore there is an element h of order p. Since p; q are distinct primes H \ K D ¹1º. Consider the element g 1 h1 gh. Since K is normal g 1 hg 2 K. Then g1 h1 gh D .g 1 h1 g/h 2 K. But H is also normal so h1 gh 2 H . This then implies that g 1 h1 gh D g 1 .h1 gh/ 2 H and therefore g 1 h1 gh 2 K \ H . It follows then that g 1 h1 gh D 1 or gh D hg. Since g; h commute the order of gh is the lcm of the orders of g and h which is pq. Therefore G has an element of order pq. Since jGj D pq this implies that G is cyclic. In the above theorem since we assumed that q is not congruent to 1 mod p and hence p ¤ 2. In the case when p D 2 we get another possibility. Theorem 13.4.4. Let p be an odd prime and G a finite group of order 2p. Then either G is cyclic or G is isomorphic to the dihedral group of order 2p, that is the group of symmetries of a regular p-gon. In this latter case G is generated by two elements g and h which satisfy the relations g p D h2 D .gh/2 D 1. Proof. As in the proof of Theorem 13.4.3 G must have a normal cyclic subgroup of order p say hgi. Since 2jjGj the group G must have an element of order 2 say h. Consider the order of .gh/. By Lagrange’s theorem this element can have order 1; 2; p; 2p. If the order is 1 then gh D 1 or g D h1 D h. This is impossible

189

Section 13.4 Some Applications of the Sylow Theorems

since g has order p and h has order 2. If the order of gh is p then from the second Sylow theorem gh 2 hgi. But this implies that h 2 hgi which is impossible since every nontrivial element of hgi has order p. Therefore the order of gh is either 2 or 2p. If the order of gh is 2p then since G has order 2p it must be cyclic. If the order of gh is 2 then within G we have the relations gp D h2 D .gh/2 D 1. Let H D hg; hi be the subgroup of G generated by g and h. The relations g p D h2 D .gh/2 D 1 imply that H has order 2p. Since jGj D 2p we get that H D G. G is isomorphic to the dihedral group Dp of order 2p (see exercises). In the above description g represents a rotation of 2 p of a regular p-gon about its center while h represents any reflection across a line of symmetry of the regular p-gon. We have looked at the finite fields Zp . We give an example of a p-Sylow subgroup of a matrix group over Zp . Example 13.4.5. Consider GL.n; p/, the group of n  n invertible matrices over Zp . If ¹v1 ; : : : ; vn º is a basis for .Zp /n over Zp then the size of GL.n; p/ is the number of independent images ¹w1 ; : : : ; wn º of ¹v1 ; : : : ; vn º. For w1 there are p n  1 choices, for w2 there are p n  p choices and so on. It follows that jGL.n; p/j D .p n  1/.p n  p/    .p n  p n1 / D p 1C2CC.n1/ m D p

n.n1/ 2

m

n.n1/

with .p; m/ D 1. Therefore a p-Sylow subgroup must have size p 2 . Let P be the subgroup of upper triangular matrices with 1’s on the diagonal. n.n1/ Then P has size p 1C2CC.n1/ D p 2 and is therefore a p-Sylow subgroup of GL.n; p/. The final example is a bit more difficult. We mentioned that a major result on finite groups is the classification of the finite simple groups. This classification showed that any finite simple group is either cyclic of prime order, in one of several classes of groups such as the An or one of a number of special examples called sporadic groups. One of the major tools in this classification is the following famous result called the Feit–Thompson theorem that showed that any finite group G of odd order is solvable and, in addition, if G is not cyclic then G is nonsimple. Theorem 13.4.6 (Feit–Thompson theorem). Any finite group of odd order is solvable. The proof of this theorem, one of the major results in algebra in the twentieth century is way beyond the scope of this book – the proof is actually hundreds of pages in length when one counts the results used. However we look at the smallest nonabelian simple group.

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Chapter 13 Groups Actions and the Sylow Theorems

Theorem 13.4.7. Suppose that G is a simple group of order 60. Then G is isomorphic to A5 . Further A5 is the smallest nonabelian finite simple group. Proof. Suppose that G is a simple group of order 60 D 22  3  5. The number of 5Sylow subgroups is of the form 1 C 5k and divides 12. Hence there is 1 or 6. Since G is assumed simple and all 5-Sylow subgroups are conjugate there cannot be only one and hence there are 6. Since each of these is cyclic of order 5 they intersect only in the identity. Hence these 6 subgroups cover 24 distinct elements. The number of 3-Sylow subgroups is of the form 1 C 3k and divides 20. Hence there are 1; 4; 10. We claim that there are 10. There can’t be only 1 since G is simple. Suppose there were 4. Let G act on the set of 3-Sylow subgroups by conjugation. Since an action is a permutation this gives a homomorphism f from G into S4 . By the first isomorphism theorem G= ker.f / Š im.f /. However since G is simple the kernel must be trivial and this implies that G would imbed into S4 . This is impossible since jGj D 60 > 24 D jS4 j. Therefore there are 10 3-Sylow subgroups. Since each of these is cyclic of order 3 they intersect only in the identity and therefore these 10 subgroups cover 20 distinct elements. Hence together with the elements in the 5-Sylow subgroups we have 44 nontrivial elements. The number of 2-Sylow subgroups is of the form 1 C 2k and divides 15. Hence there are 1; 3; 5; 15. We claim that there are 5. As before there can’t be only 1 since G is simple. There can’t be 3 since as for the case of 3-Sylow subgroups this would imply an imbedding of G into S3 which is impossible since jS3 j D 6. Suppose that there were 15 2-Sylow subgroups each of order 4. The intersections would have a maximum of 2 elements and therefore each of these would contribute at least 2 distinct elements. This gives a minimum of 30 distinct elements. However we already have 44 nontrivial elements from the 3-Sylow and 5-Sylow subgroups. Since jGj D 60 this is too many. Therefore G must have 5 2-Sylow subgroups. Now let G act on the set of 2-Sylow subgroups. This then as above implies an imbedding of G into S5 so we may consider G as a subgroup of S5 . However the only subgroup of S5 of order 60 is A5 and therefore G Š A5 . The proof that A5 is the smallest nonabelian simple group is actually brute force. We show that any group G of order less than 60 either has prime order or is nonsimple. There are strong tools that we can use. By the Feit–Thompson theorem we must only consider groups of even order. From Theorem 13.4.4 we don’t have to consider orders 2p. The rest can be done by an analysis using Sylow theory. For example we show that any group of order 20 is nonsimple. Since 20 D 22  5 the number of 5-Sylow subgroups is 1 C 5k and divides 4. Hence there is only one and therefore it must be normal and so G is nonsimple. There is a strong theorem, whose proof is usually done with representation theory, which says that any group whose order is divisible by only two primes is solvable. Therefore for jGj D 60 we only have to show that groups of order 30 D 2  3  5 and 42 D 2  3  7 are nonsimple. This is done in the same manner

Section 13.5 Exercises

191

as the first part of this proof. Suppose jGj D 30. The number of 5-Sylow subgroups is of the form 1 C 5k and divides 6 hence there are 1 or 6. If G were simple there would have to be 6 covering 24 distinct elements. The number of 3-Sylow subgroups is of the form 1 C 3k and divides 10 hence there are 1; 4; 10. If there were 10 these would cover an additional 20 distinct elements which is impossible since we already have 24 and G has order 30. If there were 4 and G were simple then G would imbed into S4 , again impossible since jGj D 30 > 24. Therefore there is only one and hence a normal 3-Sylow subgroup. It follows that G cannot be simple. The case jGj D 42 is even simpler. There must be a normal 7-Sylow subgroup.

13.5

Exercises

1. Prove Lemma 13.1.3. 2. Let the group G act on itself by conjugation. That is g.g1 / D g 1 g1 g. Prove that this is an action on the set G. 3. Show that the dihedral group Dn of order 2n has the presentation hr; f I r n D f 2 D .rf /2 D 1i. 4. Show that each group of order  59 is solvable. 5. Let P1 and P2 be two different p-Sylow subgroups of a finite group G. Show that P1 P2 is not a subgroup of G. 6. Let G be a finite group. For a prime p the following are equivalent: (i) G has exactly one p-Sylow subgroup. (ii) The product of two elements of order p has again order p. 7. Let P and Q be two p-Sylow subgroups of the finite group G. If Z.P / is a normal subgroup of Q, then Z.P / D Z.Q/.   8. Let p be a prime and G D SL.2; p/. Let P D hai, where a D 10 11 . (i) Determine the normalizer NG .P / and the number of p-Sylow subgroups of G. (ii) Determine the centralizer CG .a/. How many elements of order p does G have? In how many conjugacy classes can they be decomposed? (iii) Show that all subgroups of G of order p.p  1/ are conjugate. (iv) Show that G has no elements of order p.p  1/ for p  5.

Chapter 14

Free Groups and Group Presentations

14.1

Group Presentations and Combinatorial Group Theory

In discussing the symmetric group on 3 symbols and then the various dihedral groups in Chapters 9, 10 and 11 we came across the concept of a group presentation. Roughly for a group G a presentation consists of a set of generators X for G, so that G D hXi, and a set of relations between the elements of X from which in principle the whole group table can be constructed. In this chapter we make this concept precise. As we will see, every group G has a presentation but it is mainly in the case where the group is finite or countably infinite that presentations are most useful. Historically the idea of group presentations arose out of the attempt to describe the countably infinite fundamental groups that came out of low dimensional topology. The study of groups using group presentations is called combinatorial group theory. Before looking at group presentations in general we revisit two examples of finite groups and then a class of infinite groups. Consider the symmetric group on 3 symbols, S3 . We saw that it has the following 6 elements:       1 2 3 1 2 3 1 2 3 1D ; aD ; bD 1 2 3 2 3 1 3 1 2       1 2 3 1 2 3 1 2 3 cD ; dD ; eD : 2 1 3 3 2 1 1 3 2 Notice that a 3 D 1, c 2 D 1 and that ac D ca2 . We claim that ha; cI a3 D c 2 D .ac/2 D 1i is a presentation for S3 . First it is easy to show that S3 D ha; ci. Indeed 1 D 1;

a D a;

b D a2 ;

c D c;

d D ac;

e D a2 c

and so clearly a; c generate S3 . Now from .ac/2 D acac D 1 we get that ca D a2 c. This implies that if we write any sequence (or word in our later language) in a and c we can also rearrange it so that the only powers of a are a and a2 , the only powers of c are c and all a terms precede c terms. For example aca2 cac D aca.acac/ D a.ca/ D a.a2 c/ D .a3 /c D c:

193

Section 14.2 Free Groups

Therefore using the three relations form the presentation above each element of S3 can be written as a˛ c ˇ with ˛ D 0; 1; 2 and ˇ D 0; 1. From this the multiplication of any two elements can be determined. Exactly this type of argument applies to all the dihedral groups Dn . We saw that in general jDn j D 2n. Since these are the symmetry groups of a regular n-gon we always have a rotation r of angle 2 n about the center of the n-gon. This element r would have order n. Let f be a reflection about any line of symmetry. Then f 2 D 1 and rf is a reflection about the rotated line which is also a line of symmetry. Therefore .rf /2 D 1. Exactly as for S3 the relation .rf /2 D 1 implies that f r D r 1 f D r n1 f . This allows us to always place r terms in front of f terms in any word on r and f . Therefore the elements of Dn are always of the form r˛f ˇ ;

˛ D 0; 1; 2; : : : ; n  1;

ˇ D 0; 1

and further the relations r n D f 2 D .rf /2 D 1 allow us to rearrange any word in r and f into this form. It follows that jhr; f ij D 2n and hence Dn D hr; f i together with the relations above. Hence we obtain: Theorem 14.1.1. If Dn is the symmetry group of a regular n-gon then a presentation for Dn is given by Dn D hr; f I r n D f 2 D .rf /2 D 1i: We now give one class of infinite examples. If G is an infinite cyclic group, so that G Š Z, then G D hgI i is a presentation for G. That is G has a single generator with no relations. A direct product of n copies of Z is called a free abelian group of rank n. We will denote this by Zn . A presentation for Zn is then given by Zn D hx1 ; x2 ; : : : ; xn I xi xj D xj xi for all i; j D 1; : : : ; ni:

14.2

Free Groups

Crucial to the concept of a group presentation is the idea of a free group. Definition 14.2.1. A group F is free on a subset X if every map f W X ! G with G a group can be extended to a unique homomorphism f W F ! G. X is called a free basis for F . In general a group F is a free group if it is free on some subset X . If X is a free basis for a free group F we write F D F .X /. We first show that given any set X there does exist a free group with free basis X. Let X D ¹xi ºi2I be a set (possibly empty). We will construct a group F .X / which is free with free basis X. First let X 1 be a set disjoint from X but bijective to X. If xi 2 X then the corresponding element of X 1 under the bijection we denote xi1 and

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Chapter 14 Free Groups and Group Presentations

say that xi and xi1 are associated. The set X 1 is called the set of formal inverses from X and we call X [X 1 the alphabet. Elements of the alphabet are called letters, hence a letter has the form xi1 where i D ˙1. A word in X is a finite sequence of letters from the alphabet. That is a word has the form i

i



w D xi1 1 xi22 : : : xinin where xij 2 X and ij D ˙1. If n D 0 we call it the empty word which we will denote e. The integer n is called the length of the word. Words of the form xi xi1 or xi1 xi are called trivial words. We let W .X / be the set of all words on X. If w1 ; w2 2 W .X / we say that w1 is equivalent to w2 , denoted w1  w2 , if w1 can be converted to w2 by a finite string of insertions and deletions of trivial words. For example if w1 D x3 x4 x41 x2 x2 and w2 D x3 x2 x2 then w1  w2 . It is straightforward to verify that this is an equivalence relation on W .X / (see exercises). Let F .X / denote the set of equivalence classes in W .X / under this relation, hence F .X / is a set of equivalence classes of words from X. A word w 2 W .X / is said to be freely reduced or reduced if it has no trivial subwords (a subword is a connected sequence within a word). Hence in the example above w2 D x3 x2 x2 is reduced but w1 D x3 x4 x41 x2 x2 is not reduced. In each equivalence class in F .X / there is a unique element of minimal length. Further this element must be reduced or else it would be equivalent to something of smaller length. Two reduced words in W .X / are either equal or not in the same equivalence class in F .X /. Hence F .X / can also be considered as the set of all reduced words from W .X /. i i  Given a word w D xi1 1 xi22 : : : xinin we can find the unique reduced word w equivalent to w via the following free reduction process. Beginning from the left side of w we cancel each occurrence of a trivial subword. After all these possible cancellations we have a word w 0 . Now we repeat the process again starting from the left side. Since w has finite length eventually the resulting word will either be empty or reduced. The final reduced w is the free reduction of w. Now we build a multiplication on F .X /. If i

i



w1 D xi1 1 xi22 : : : xinin ;

j

j



jm w2 D xj1 1 xj2 2 : : : xjm

are two words in W .X / then their concatenation w1 ? w2 is simply placing w2 after w1 , i i j j  jm : w1 ? w2 D xi1 1 xi22 : : : xinin xj1 1 xj2 2 : : : xjm If w1 ; w2 2 F .x/ then we define their product as w1 w2 D equivalence class of w1 ? w2 : That is we concatenate w1 and w2 , and the product is the equivalence class of the resulting word. It is easy to show that if w1  w10 and w2  w20 then w1 ? w2 

195

Section 14.2 Free Groups

w10 ? w20 so that the above multiplication is well-defined. Equivalently we can think of this product in the following way. If w1 ; w2 are reduced words then to find w1 w2 first j  concatenate and then freely reduce. Notice that if xinin xj1 1 is a trivial word then it is cancelled when the concatenation is formed. We say then that there is cancellation in forming the product w1 w2 . Otherwise the product is formed without cancellation. Theorem 14.2.2. Let X be a nonempty set and let F .X / be as above. Then F .X / is a free group with free basis X. Further if X D ; then F .X / D ¹1º, if jX j D 1 then F .X / Š Z and if jX j  2 then F .X / is nonabelian. Proof. We first show that F .X / is a group and then show that it satisfies the universal mapping property on X. We consider F .X / as the set of reduced words in W .X / with the multiplication defined above. Clearly the empty word acts as the identity i i i i   element 1. If w D xi11 xi22 : : : xinin and w1 D xin in xin1n1 : : : xi1 1 then both w ? w1 and w1 ? w freely reduce to the empty word and so w1 is the inverse of w. Therefore each element of F .X / has an inverse. Therefore to show that F .X / forms a group we must show that the multiplication is associative. Let i

i



w1 D xi1 1 xi22 : : : xinin ;

j

j



jm w2 D xj1 1 xj2 2 : : : xjm ;

k

k

k

w3 D xk11 xk22 : : : xkpp

be three freely reduced words in F .X /. We must show that .w1 w2 /w3 D w1 .w2 w3 /: To prove this we use induction on m, the length of w2 . If m D 0 then w2 is the empty word and hence the identity and it is certainly true. Now suppose that m D 1 j so that w2 D xj1 1 . We must consider exactly four cases. j

Case (1): There is no cancellation in forming either w1 w2 or w2 w3 . That is xj1 1 ¤



j

k

xin in and xj1 1 ¤ xk1 1 . Then the product w1 w2 is just the concatenation of the words and so is .w1 w2 /w3 . The same is true for w1 .w2 w3 /. Therefore in this case w1 .w2 w3 / D .w1 w2 /w3 . Case (2): There is cancellation in forming w1 w2 but not in forming w2 w3 . Then if we concatenate all three words the only cancellation occurs between w1 and w2 in either w1 .w2 w3 / or in .w1 w2 /w3 and hence they are equal. Therefore in this case w1 .w2 w3 / D .w1 w2 /w3 . Case (3): There is cancellation in forming w2 w3 but not in forming w1 w2 . This is entirely analogous to Case (2) so therefore in this case w1 .w2 w3 / D .w1 w2 /w3 . Case (4): There is cancellation in forming w1 w2 and also in forming w2 w3 . Then j j k  xj1 1 D xin in and xj1 1 D xk1 1 . Here i

i

k

k

k

n1 xk11 xk22 : : : xkpp : .w1 w2 /w3 D xi1 1 : : : xin1

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Chapter 14 Free Groups and Group Presentations

On the other hand i



k

k

w1 .w2 w3 / D xi11 : : : xinin xk22 : : : xkpp : 

k

However these are equal since xinin D xk11 . Therefore in this final case w1 .w2 w3 / D .w1 w2 /w3 . It follows inductively from these four cases that the associative law holds in F .X / and therefore F .X / forms a group. Now suppose that f W X ! G is a map from X into a group G. By the construction of F .X / as a set of reduced words this can be extended to a unique homomorphism. i  If w 2 F with w D xi11 : : : xinin then define f .w/ D f .xi1 /i1    f .xin /in . Since multiplication in F .X / is concatenation this defines a homomorphism and again form the construction of F .X / its the only one extending f . This is analogous to constructing a linear transformation from one vector space to another by specifying the images of a basis. Therefore F .X / satisfies the universal mapping property of Definition 14.2.1 and hence F .X / is a free group with free basis X. The final parts of Theorem 14.2.2 are straightforward. If X is empty the only reduced word is the empty word and hence the group is just the identity. If X has a single letter then F .X / has a single generator and is therefore cyclic. It is easy to see that it must be torsion-free and therefore F .X / is infinite cyclic, that is F .x/ Š Z. Finally if jX j  2 let x1 ; x2 2 X. Then x1 x2 ¤ x2 x1 and both are reduced. Therefore F .X / is nonabelian. The proof of Theorem 14.2.2 provides another way to look at free groups. Theorem 14.2.3. F is a free group if and only if there is a generating set X such that every element of F has a unique representation as a freely reduced word on X. The structure of a free group is entirely dependent on the cardinality of a free basis. In particular the cardinality of a free basis jX j for a free group F is unique and is called the rank of F . If jX j < 1, F is of finite rank. If F has rank n and X D ¹x1 ; : : : ; xn º we say that F is free on ¹x1 ; : : : ; xn º. We denote this by F .x1 ; x2 ; : : : ; xn /. Theorem 14.2.4. If X and Y are sets with the same cardinality, that is jX j D jY j, then F .X / Š F .Y /, the resulting free groups are isomorphic. Further if F .X / Š F .Y / then jX j D jY j. Proof. Suppose that f W X ! Y is a bijection from X onto Y . Now Y  F .Y / so there is a unique homomorphism  W F .X / ! F .Y / extending f . Since f is a bijection it has an inverse f 1 W Y ! X and since F .Y / is free there is a unique homomorphism 1 from F .Y / to F .X / extending f 1 . Then 1 is the identity map on F .Y / and 1  is the identity map on F .X /. Therefore ; 1 are isomorphisms with  D 11 .

197

Section 14.2 Free Groups

Conversely suppose that F .X / Š F .Y /. In F .X / let N.X / be the subgroup generated by all squares in F .X / that is N.X / D h¹g2 W g 2 F .X /ºi: Then N.X / is a normal subgroup and the factor group F .X /=N.X / is abelian where every nontrivial element has order 2 (see exercises). Therefore F .X /=N.X / can be considered as a vector space over Z2 , the finite field of order 2, with X as a vector space basis. Hence jX j is the dimension of this vector space. Let N.Y / be the corresponding subgroup of F .Y /. Since F .X / Š F .Y / we would have F .X /=N.X / Š F .Y /=N.Y / and therefore jY j is the dimension of the vector space F .Y /=N.Y /. Therefore jX j D jY j from the uniqueness of dimension of vector spaces. Expressing elements of F .X / as a reduced word gives a normal form for elements in a free group F . As we will see in Section 14.5 this solves what is termed the word problem for free groups. Another important concept is the following: a freely reduced word W D xve11 xve22 : : : xvenn is cyclically reduced if v1 ¤ vn or if v1 D vn then e1 ¤ en . Clearly then every element of a free group is conjugate to an element given by a cyclically reduced word. This provides a method to determine conjugacy in free groups. Theorem 14.2.5. In a free group F two elements g1 ; g2 are conjugate if and only if a cyclically reduced word for g1 is a cyclic permutation of a cyclically reduced word for g2 . The theory of free groups has a large and extensive literature. We close this section by stating several important properties. Proofs for these results can be found in [24], [23] or [15]. Theorem 14.2.6. A free group is torsion-free. From Theorem 14.2.4 we can deduce: Theorem 14.2.7. An abelian subgroup of a free group must be cyclic. Finally a celebrated theorem of Nielsen and Schreier states that a subgroup of a free group must be free. Theorem 14.2.8 (Nielsen–Schreier). A subgroup of a free group is itself a free group. Combinatorially F is free on X if X is a set of generators for F and there are no nontrivial relations. In particular:

198

Chapter 14 Free Groups and Group Presentations

There are several different proofs of this result (see [24]) with the most straightforward being topological in nature. We give an outline of a simple topological proof in Section 14.4. Nielsen, using a technique now called Nielsen transformations in his honor, first proved this theorem about 1920 for finitely generated subgroups. Schreier shortly after found a combinatorial method to extend this to arbitrary subgroups. A complete version of the original combinatorial proof appears in [24] and in the notes by Johnson [19]. Schreier’s combinatorial proof also allows for a description of the free basis for the subgroup. In particular, let F be free on X, and H  F a subgroup. Let T D ¹t˛ º be a complete set of right coset representatives for F mod H with the property that if t˛ D xve11 xve22 : : : xvenn 2 T , with i D ˙1 then all the initial segments 1; xve11 ; xve11 xve22 , etc. are also in T . Such a system of coset representatives can always be found and is called a Schreier system or Schreier transversal for H . If g 2 F let g represent its coset representative in T and further define for g 2 F and t 2 T , S tg D tg.tg/1 . Notice that S tg 2 H for all t; g. We then have: Theorem 14.2.9 (explicit form of Nielsen–Schreier). Let F be free on X and H a subgroup of F . If T is a Schreier transversal for F mod H then H is free on the set ¹Stx W t 2 T; x 2 X; S tx ¤ 1º. Example 14.2.10. Let F be free on ¹a; bº and H D F .X 2 / the normal subgroup of F generated by all squares in F . Then F=F .X 2 / D ha; bI a2 D b 2 D .ab/2 D 1i D Z2  Z2 . It follows that a Schreier system for F mod H is ¹1; a; b; abº with a D a; b D b and ba D ab. From this it can be shown that H is free on the generating set x1 D a 2 ;

x2 D bab 1 a1 ;

x3 D b 2 ;

x4 D abab 1 ;

x5 D ab 2 a1 :

The theorem also allows for a computation of the rank of H given the rank of F and the index. Specifically: Corollary 14.2.11. Suppose F is free of rank n and jF W H j D k. Then H is free of rank nk  k C 1. From the example we see that F is free of rank 2, H has index 4 so H is free of rank 2  4  4 C 1 D 5.

14.3

Group Presentations

The significance of free groups stems from the following result which is easily deduced from the definition and will lead us directly to a formal definition of a group

Section 14.3 Group Presentations

199

presentation. Let G be any group and F the free group on the elements of G considered as a set. The identity map f W G ! G can be extended to a homomorphism of F onto G, therefore: Theorem 14.3.1. Every group G is a homomorphic image of a free group. That is let G be any group. Then G D F=N where F is a free group. In the above theorem instead of taking all the elements of G we can consider just a set X of generators for G. Then G is a factor group of F .X /, G Š F .X /=R. The normal subgroup N is the kernel of the homomorphism from F .X / onto G. We use Theorem 14.3.1 to formally define a group presentation. If H is a subgroup of a group G then the normal closure of H denoted by N.H / is the smallest normal subgroup of G containing H . This can be described alternatively in the following manner. The normal closure of H is the subgroup of G generated by all conjugates of elements of H . Now suppose that G is a group with X a set of generators for G. We also call X a generating system for G. Now let G D F .X /=N as in Theorem 14.3.1 and the comments after it. N is the kernel of the homomorphism f W F .X / ! G. It follows that if r is a free group word with r 2 N then r D 1 in G (under the homomorphism). We then call r a relator in G and the equation r D 1 a relation in G. Suppose that R is a subset of N such that N D N.R/, then R is called a set of defining relators for G. The equations r D 1, r 2 R, are a set of defining relations for G. It follows that any relator in G is a product of conjugates of elements of R. Equivalently r 2 F .X / is a relator in G if and only if r can be reduced to the empty word by insertions and deletions of elements of R and trivial words. Definition 14.3.2. Let G be a group. Then a group presentation for G consists of a set of generators X for G and a set R of defining relators. In this case we write G D hXI Ri. We could also write the presentation in terms of defining relations as G D hXI r D 1; r 2 Ri. From Theorem 14.3.1 it follows immediately that every group has a presentation. However in general there are many presentations for the same group. If R  R1 then R1 is also a set of defining relators. Lemma 14.3.3. Let G be a group. Then G has a presentation. If G D hXI Ri and X is finite then G is said to be finitely generated. If R is finite G is finitely related. If both X and R are finite G is finitely presented. Using group presentations we get another characterization of free groups. Theorem 14.3.4. F is a free group if and only if F has a presentation of the form F D hXI i.

200

Chapter 14 Free Groups and Group Presentations

Mimicking the construction of a free group from a set X we can show that to each presentation corresponds a group. Suppose that we are given a supposed presentation hXI Ri where R is given as a set of words in X. Consider the free group F .X / on X. Define two words w1 ; w2 on X to be equivalent if w1 can be transformed into w2 using insertions and deletions of elements of R and trivial words. As in the free group case this is an equivalence relation. Let G be the set of equivalence classes. If we define multiplication as before as concatenation followed by the appropriate equivalence class then G is a group. Further each r 2 R must equal the identity in G so that G D hX I Ri. Notice that here there may be no unique reduced word for an element of G. Theorem 14.3.5. Given hXI Ri where X is a set and R is a set of words on X. Then there exists a group G with presentation hXI Ri. We now give some examples of group presentations. Example 14.3.6. A free group of rank n has a presentation Fn D hx1 ; : : : ; xn I i: Example 14.3.7. A free abelian group of rank n has a presentation Zn D hx1 ; : : : ; xn I xi xj xi1 xj1 ; i D 1; : : : ; n; j D 1; : : : ; ni: Example 14.3.8. A cyclic group of order n has a presentation Zn D hxI x n D 1i: Example 14.3.9. The dihedral groups of order 2n representing the symmetry group of a regular n-gon has a presentation hr; f I r n D 1; f 2 D 1; .rf /2 D 1i: We look at this example in Section 14.3.1.

14.3.1 The Modular Group In this section we give a more complicated example and then a nice application to number theory. If R is any commutative ring with an identity, then the set of invertible nn matrices with entries from R forms a group under matrix multiplication called the n-dimensional general linear group over R (see [25]). This group is denoted by GLn .R/. Since det.A/ det.B/ D det.AB/ for square matrices A; B it follows that the subset of GLn .R/ consisting of those matrices of determinant 1 forms a subgroup. This

Section 14.3 Group Presentations

201

subgroup is called the special linear group over R and is denoted by SLn .R/. In this section we concentrate on SL2 .Z/, or more specifically a quotient of it, PSL2 .Z/ and find presentations for them. The group SL2 .Z/ then consists of 2  2 integral matrices of determinant one: ²  ³ a b W a; b; c; d 2 Z; ad  bc D 1 : SL2 .Z/ D c d SL2 .Z/ is called the homogeneous modular group and an element of SL2 .Z/ is called a unimodular matrix. If G is any group, recall that its center Z.G/ consists of those elements of G which commute with all elements of G: Z.G/ D ¹g 2 G W gh D hg; 8h 2 Gº: Z.G/ is a normal subgroup of G and hence we can form the factor group G=Z.G/. For G D SL2 .Z/ the only unimodular matrices that commute with all others are ˙I D ˙ 10 01 . Therefore Z.SL2 .Z// D ¹I; I º. The quotient SL2 .Z/=Z.SL2 .Z// D SL2 .Z/=¹I; I º is denoted PSL2 .Z/ and is called the projective special linear group or inhomogeneous modular group. More commonly PSL2 .Z/ is just called the Modular Group and denoted by M . M arises in many different areas of mathematics including number theory, complex analysis and Riemann surface theory and the theory of automorphic forms and functions. M is perhaps the most widely studied single finitely presented group. Complete discussions of M and its structure can be found in the books Integral Matrices by M. Newman [38] and Algebraic Theory of the Bianchi Groups by B. Fine [34]. Since M D PSL2 .Z/ D SL2 .Z/=¹I; I º it follows that each element of M can be considered as ˙A where A is a unimodular matrix. A projective unimodular matrix is then   a b ˙ ; a; b; c; d 2 Z; ad  bc D 1: c d The elements of M can also be considered as linear fractional transformations over the complex numbers z0 D

az C b ; cz C d

a; b; c; d 2 Z; ad  bc D 1; where z 2 C:

Thought of in this way, M forms a Fuchsian group which is a discrete group of isometries of the non-Euclidean hyperbolic plane. The book by Katok [20] gives a solid and clear introduction to such groups. This material can also be found in condensed form in [35].

202

Chapter 14 Free Groups and Group Presentations

We now determine presentations for both SL2 .Z/ and M D PSL2 .Z/. Theorem 14.3.10. The group SL2 .Z/ is generated by the elements     0 1 0 1 XD and Y D : 1 0 1 1 Further a complete set of defining relations for the group in terms of these generators is given by X 4 D Y 3 D YX 2 Y 1 X 2 D I: It follows that SL2 .Z/ has the presentation hX; Y I X 4 D Y 3 D YX 2 Y 1 X 2 D I i: Proof. We first show that SL2 .Z/ is generated by X and Y , that is every matrix A in the group can be written as a product of powers of X and Y . Let   1 1 U D : 0 1 Then a direct multiplication shows that U D X Y and we show that SL2 .Z/ is generated by X and U which implies that it is also generated by X and Y . Further   1 n n U D 0 1 so that U has order.   infinite Let A D ac db 2 SL2 .Z/. Then we have     a C kc b C kd c d XA D and U k A D c d a b for any k 2 Z. We may assume that jcj  jaj otherwise start with XA rather than A. If c D 0 then A D ˙U q for some q. If A D U q then certainly A is in the group generated by X and U . If A D U q then A D X 2 U q since X 2 D I . It follows that here also A is in the group generated by X and U . Now suppose c ¤ 0. Apply the Euclidean algorithm to a and c in the following modified way: a D q0 c C r1 c D q1 r1 C r2 r1 D q2 r2 C r3 :: : .1/n rn1 D qn rn C 0

203

Section 14.3 Group Presentations

where rn D ˙1 since .a; c/ D 1. Then X U qn    X U q0 A D ˙U qnC1

with qnC1 2 Z:

Therefore A D X m U q0 X U q1    X U qn X U qnC1 with m D 0; 1; 2; 3; q0 ; q1 ; : : : ; qnC1 2 Z and q0 ; : : : ; qn ¤ 0. Therefore X and U and hence X and Y generate SL2 .Z/. We must now show that X 4 D Y 3 D YX 2 Y 1 X 2 D I are a complete set of defining relations for SL2 .Z/ or that every relation on these generators is derivable from these. It is straightforward to see that X and Y do satisfy these relations. Assume then that we have a relation S D X 1 Y ˛1 X 2 Y ˛2    Y ˛n X nC1 D I with all i ; ˛j 2 Z. Using the set of relations X 4 D Y 3 D YX 2 Y 1 X 2 D I we may transform S so that S D X 1 Y ˛1    Y ˛m X mC1 with 1 ; mC1 D 0; 1; 2 or 3 and ˛i D 1 or 2 for i D 1; : : : ; m and m  0. Multiplying by a suitable power of X we obtain Y ˛1 X    Y ˛m X D X ˛ D S1 with m  0 and ˛ D 0; 1; 2 or 3. Assume that m  1 and let   a b S1 D : c d We show by induction that a; b; c; d  0;

bCc >0

a; b; c; d  0;

b C c < 0:

or This claim for the entries of S1 is true for   1 0 YX D and 1 1

Y 2X D

  1 1 : 0 1

204

Chapter 14 Free Groups and Group Presentations

Suppose it is correct for S2 D



a1 b1  . c1 d1

Then



 b1 a1 YXS2 D .a1 C c1 / b1 C d1   a1  c1 b1 C d1 2 : Y XS2 D c1 d1

and

Therefore the claim is correct for all S1 with m  1. This gives a contradiction, for the entries of X ˛ with ˛ D 0; 1; 2 or 3 do not satisfy the claim. Hence m D 0 and S can be reduced to a trivial relation by the given set of relations. Therefore they are a complete set of defining relations and the theorem is proved. Corollary 14.3.11. The modular group M D PSL2 .Z/ has the presentation M D hx; yI x 2 D y 3 D 1i: Further x; y can be taken as the linear fractional transformations x W z0 D 

1 z

and

y W z0 D 

1 : zC1

Proof. The center of SL2 .Z/ is ˙I . Since X 2 D I setting X 2 D I in the presentation for SL2 .Z/ gives the presentation for M . Writing the projective matrices as linear fractional transformations gives the second statement. This corollary says that M is the free product of a cyclic group of order 2 and a cyclic group of order 3 a concept we will introduce in Section 14.7. We note that there is an elementary alternative proof to Corollary 14.3.11 as far as showing that X 2 D Y 3 D 1 are a complete set of defining relations. As linear fractional transformations we have 1 X.z/ D  ; z Now let Then

Y .z/ D 

RC D ¹x 2 R W x > 0º X.R /  RC

and

1 ; zC1

and

Y 2 .z/ D 

zC1 : z

R D ¹x 2 R W x < 0º:

Y ˛ .RC /  R ;

˛ D 1; 2:

Let S 2 . Using the relations X 2 D Y 3 D 1 and a suitable conjugation we may assume that either S D 1 is a consequence of these relations or that S D Y ˛1 X Y ˛2    X Y ˛n with 1  ˛i  2 and ˛1 D ˛n .

205

Section 14.3 Group Presentations

In this second case if x 2 RC then S.x/ 2 R and hence S ¤ 1. This type of ping-pong argument can be used in many examples (see [23], [15] and [19]). As another example consider the unimodular matrices     0 1 0 1 AD ; BD : 1 2 1 2 Let A; B denote the corresponding linear fractional transformations in the modular group M . We have     n C 1 n n C 1 n n n A D ; B D for n 2 Z: n n C 1 n nC1 In particular A and B have infinite order. Now n

A .R /  RC

n

and B .RC /  R

for all n ¤ 0. The ping-pong argument used for any element of the type n1

S DA B

m1

B

mk

A

nkC1

with all ni ; mi ¤ 0 and n1 C nkC1 ¤ 0 shows that S.x/ 2 RC if x 2 R . It follows that there are no nontrivial relations on A and B and therefore the subgroup of M generated by A; B must be a free group of rank 2. To close this section we give a nice number theoretical application of the modular group. First we need the following corollary to Corollary 14.3.11. Corollary 14.3.12. Let M D hX; Y I X 2 D Y 3 D 1i be the modular group. If A is an element of order 2 then A is conjugate to X. If B is an element of order 3 then B is conjugate to either Y or Y 2 . Definition 14.3.13. Let a; n be relatively prime integers with a ¤ 0; n  1. Then a is a quadratic residue mod n if there exists an x 2 Z with x 2 D a mod n, that is a D x 2 C k n for some k 2 Z. The following is called Fermat’s two-square theorem. Theorem 14.3.14 (Fermat’s two-square theorem). Let n > 0 be a natural number. Then n D a2 C b 2 with .a; b/ D 1 if and only if 1 is a quadratic residue modulo n. Proof. Suppose 1 is a quadratic residue mod n. Then there exists an x with x 2  1 mod n or x 2 D 1 C mn. This implies that x 2  mn D 1 so that there must exist a projective unimodular matrix   x n AD˙ : m x

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Chapter 14 Free Groups and Group Presentations

It is straightforward that A2 D 1 so by Corollary 14.3.12 A is conjugate within M to X. Now consider conjugates of X within M . Let T D ac db . Then   d b T 1 D c a and TX T 1 D



a b c d

    0 1 d b .bd C ac/ a2 C b 2 : D˙ .c 2 C d 2 / bd C ac 1 0 c a



( )

Therefore any conjugate of X must have form ( ) and therefore A must have form ( ). Therefore n D a2 Cb 2 . Further .a; b/ D 1 since in finding form ( ) we had ad bc D 1. Conversely suppose n D a2 C b 2 with .a; b/ D 1. Then there exist c; d 2 Z with ad  bc D 1 and hence there exists a projective unimodular matrix   a b T D˙ : c d Then TX T

1

    ˛ n ˛ a2 C b 2 D˙ : D˙

˛

˛

This has determinant one, so ˛ 2  n D 1 H) ˛ 2 D 1  n H) ˛ 2  1

mod n:

Therefore 1 is a quadratic residue mod n. This type of group theoretical proof can be extended in several directions. KernIsberner and Rosenberger [21] considered groups of matrices of the form p   a b N ; a; b; c; d; N 2 Z; ad  N bc D 1 U D p c N d or   p a N p b ; a; b; c; d; N 2 Z; Nad  bc D 1: U D c d N They then proved that if N 2 ¹1; 2; 4; 5; 6; 8; 9; 10; 12; 13; 16; 18; 22; 25; 28; 37; 58º and n 2 N with .n; N / D 1 then: (1) If N is a quadratic residue mod n and n is a quadratic residue mod N then n can be written as n D x 2 C Ny 2 with x; y 2 Z. (2) Conversely if n D x 2 C Ny 2 with x; y 2 Z and .x; y/ D 1 then N is a quadratic residue mod n and n is a quadratic residue mod N . p The proof of the above results depends on the class number of Q. N / (see [21]). In another direction Fine [33] and [32] showed that the Fermat two-square property is actually a property satisfied by many rings R. These are called sum of squares rings. For example if p  3 mod 4 then Zp n for n > 1 is a sum of squares ring.

207

Section 14.4 Presentations of Subgroups

14.4

Presentations of Subgroups

Given a group presentation G D hXI Ri it is possible to find a presentation for a subgroup H of G. The procedure to do this is called the Reidemeister–Schreier process and is a consequence of the explicit version of the Nielsen–Schreier theorem (Theorem 14.2.9). We give a brief description. A complete description and a verification of its correctness is found in [24] or in [15]. Let G be a group with the presentation ha1 ; : : : ; an I R1 ; : : : ; Rk i. Let H be a subgroup of G and T a Schreier system for G mod H defined analogously as above. Reidemeister–Schreier Process. Let G; H and T be as above. Then H is generated by the set ¹S tav W t 2 T; av 2 ¹a1 ; : : : ; an º; S tav ¤ 1º with a complete set of defining relations given by conjugates of the original relators rewritten in terms of the subgroup generating set. In order to actual rewrite the relators in terms of the new generators we use a mapping  on words on the generators of G called the Reidemeister rewriting process. This map is defined as follows: If e

W D ave11 ave22 : : : avjj with ei D ˙1 defines an element of H then e

 .W / D S te11;av S te22;av    S tjj;av 1

2

j

where ti is the coset representative of the initial segment of W preceding avi if ei D 1 if and ti is the representative of the initial segment of W up to and including av1 i ei D 1. The complete set of relators rewritten in terms of the subgroup generators is then given by ¹ .tRi t 1 /º with t 2 T and Ri runs over all relators in G: We present two examples; one with a finite group and then an important example with a free group which shows that a countable free group contains free subgroups of arbitrary ranks. Example 14.4.1. Let G D A4 be the alternating group on 4 symbols. Then a presentation for G is G D A4 D ha; bI a2 D b 3 D .ab/3 D 1i: Let H D A04 the commutator subgroup. We use the above method to find a presentation for H . Now G=H D A4 =A04 D ha; bI a2 D b 3 D .ab/3 D Œa; b D 1i D hbI b 3 D 1i:

208

Chapter 14 Free Groups and Group Presentations

Therefore jA4 W A04 j D 3. A Schreier system is then ¹1; b; b 2 º. The generators for A04 are then X1 D S1a D a;

X2 D Sba D bab 1 ;

X3 D Sb 2 a D b 2 ab

while the relations are 1.  .aa/ D S1a S1a D X12 2.  .baab 1 / D X22 3.  .b 2 aab 2 / D X32 4.  .bbb/ D 1 5.  .bbbbb 1 / D 1 6.  .b 2 bbbb 2 / D 1 7.  .ababab/ D S1a Sba Sb 2 a D X1 X2 X3 8.  .babababb 1 / D Sba Sb 2 a S1a D X2 X3 X1 9.  .b 2 abababb 2 / D Sb 2 a S1a Sba D X3 X1 X2 . Therefore after eliminating redundant relations and using that X3 D X1 X2 we get as a presentation for A04 , hX1 ; X2 I X12 D X22 D .X1 X2 /2 D 1i: Example 14.4.2. Let F D hx; yI i be the free group of rank 2. Let H be the commutator subgroup. Then F=H D hx; yI Œx; y D 1i D Z  Z a free abelian group of rank 2. It follows that H has infinite index in F . As Schreier coset representatives we can take tm;n D x m y n ;

m D 0; ˙1; ˙2; : : : ; n D 0; ˙1; ˙2; : : : :

The corresponding Schreier generators for H are xm;n D x m y n x m y n ;

m D 0; ˙1; ˙2; : : : ; n D 0; ˙1; ˙2; : : : :

The relations are only trivial and therefore H is free on the countable infinitely many generators above. It follows that a free group of rank 2 contains as a subgroup a free group of countably infinite rank. Since a free group of countable infinite rank contains as subgroups free groups of all finite ranks it follows that a free group of rank 2 contains as a subgroup a free subgroup of any arbitrary finite rank.

Section 14.5 Geometric Interpretation

209

Theorem 14.4.3. Let F be free of rank 2. Then the commutator subgroup F 0 is free of countable infinite rank. In particular a free group of rank 2 contains as a subgroup a free group of any finite rank n. Corollary 14.4.4. Let n; m be any pair of positive integers n; m  2 and Fn , Fm free groups of ranks n; m respectively. Then Fn can be embedded into Fm and Fm can be embedded into Fn .

14.5

Geometric Interpretation

Combinatorial group theory has its origins in topology and complex analysis. Especially important in the development is the theory of the fundamental group. This connection is so deep that many people consider combinatorial group theory as the study of the fundamental group – especially the fundamental group of a low-dimensional complex. This connection proceeds in both directions. The fundamental group provides methods and insights to study the topology. In the other direction the topology can be used to study the groups. Recall that if X is a topological space then its fundamental group based at a point x0 , denoted .X; x0 /, is the group of all homotopy classes of closed paths at x0 . If X is path connected then the fundamental groups at different points are all isomorphic and we can speak of the fundamental group of X which we will denote .X /. Historically group presentations were developed to handle the fundamental groups of spaces which allowed simplicial or cellular decompositions. In these cases the presentation of the fundamental group can be read off from the combinatorial decomposition of the space. An (abstract) simplicial complex or cell complex K is a topological space consisting of a set of points called the vertices, which we will denote V .K/, and collections of subsets of vertices called simplexes or cells which have the property that the intersection of any two simplices is again a simplex. If n is the number of vertices in a cell then n  1 is called its dimension. Hence the set of vertices are the 0-dimensional cells and a simplex ¹v1 ; : : : ; vn º is an .n  1/-dimensional cell. The 1-dimensional cells are called edges. These have the form ¹u; vº where u and v are vertices. One should think of the cells in a geometric manner so that the edges are really edges, the 2-cells are filled triangles, that are equivalent to disks and so on. The maximum dimension of any cell in a complex K is called the dimension of K. From now on we will assume that our simplicial complexes are path connected. A graph  is just a 1-dimensional simplicial complex. Hence  consists of just vertices and edges. If K is any complex then the set of vertices and edges is called the 1-skeleton of K. Similarly all the cells of dimension less than or equal to 2 comprise the 2-skeleton. A connected graph with no closed paths in it is called a tree. If K is

210

Chapter 14 Free Groups and Group Presentations

any complex then a maximal tree in K is a tree that can be contained in no other tree within K. From the viewpoint of combinatorial group theory what is relevant is that if K is a complex then a presentation of its fundamental group can be determined from its 2-skeleton and read off directly. In particular: Theorem 14.5.1. Suppose that K is a connected cell complex. Suppose that T is a maximal tree within the 1-skeleton of K. Then a presentation for .K/ can be determined in the following manner: Generators: all edges outside of the maximal tree T Relations:

(a) ¹u; vº D 1 if ¹u; vº is an edge in T (b) ¹u; vº¹v; wº D ¹u; wº if u; v; w lie in a simplex of K.

From this the following is obvious: Corollary 14.5.2. The fundamental group of a connected graph is free. Further its rank is the number of edges outside a maximal tree. A connected graph is homotopic to a wedge or bouquet of circles. If there are n circles in a bouquet of circles then the fundamental group is free of rank n. The converse is also true. A free group can be realized as the fundamental group of a wedge of circles. An important concept in applying combinatorial group theory is that of a covering complex. Definition 14.5.3. Suppose that K is a complex. Then a complex K1 is a covering complex for K if there exists a surjection p W K1 ! K called a covering map with the property that for any cell s 2 K the inverse image p 1 .s/ is a union of pairwise disjoint cells in K1 and p restricted to any of the preimage cells is a homeomorphism. That is for each simplex S in K we have [ p 1 .S/ D Si and p W Si ! S is a bijection for each i . The following then becomes clear. Lemma 14.5.4. If K1 is a connected covering complex for K then K1 and K have the same dimension. What is crucial in using covering complexes to study the fundamental group is that there is a Galois theory of covering complexes and maps. The covering map p induces a homomorphism of the fundamental group which we will also call p. Then:

Section 14.5 Geometric Interpretation

211

Theorem 14.5.5. Let K1 be a covering complex of K with covering map p. Then p..K1 // is a subgroup of .K/. Conversely to each subgroup H of .K/ there is a covering complex K1 with .K1 / D H . Hence there is a one-to-one correspondence between subgroups of the fundamental group of a complex K and covers of K. We will see the analog of this theorem in regard to algebraic field extensions in Chapter 15. A topological space X is simply connected if .X / D ¹1º. Hence the covering complex of K corresponding to the identity in .K/ is simply connected. This is called the universal cover of K since it covers any other cover of K. Based on Theorem 14.5.1 we get a very simple proof of the Nielsen–Schreier theorem. Theorem 14.5.6 (Nielsen–Schreier). Any subgroup of a free group is free. Proof. Let F be a free group. Then F D .K/ where K is a connected graph. Let H be a subgroup of F . Then H corresponds to a cover K1 of K. But a cover is also 1-dimensional and hence H D .K1 / where K1 is a connected graph and hence H is also free. The fact that a presentation of a fundamental group of a simplicial complex is determined by its 2-skeleton goes in the other direction also. That is given an arbitrary presentation there exists a 2-dimensional complex whose fundamental group has that presentation. Essentially given a presentation hXI Ri we consider a wedge of circles with cardinality jX j. We then paste on a 2-cell for each relator W in R bounded by the path corresponding to the word W . Theorem 14.5.7. Given an arbitrary presentation hXI Ri there exists a connected 2-complex K with .K/ D hXI Ri. We note that the books by Rotman [26] and Camps, große Rebel and Rosenberger [15] have very nice detailed and accessible descriptions of groups and complexes. Cayley and then Dehn introduced for each group G, a graph, now called its Cayley graph, as a tool to apply complexes to the study of G. The Cayley graph is actually tied to a presentation and not to the group itself. Gromov reversed the procedure and showed that by considering the geometry of the Cayley graph one could get information about the group. This led to the development of the theory of hyperbolic groups (see for instance [15]). Definition 14.5.8. Let G D hXI Ri be a presentation. We form a graph .G; X / in the following way. Let A D X [ X 1 . For the vertex set of .G; X / we take the elements of G, that is V ./ D ¹g W g 2 Gº. The edges of  are given by the set ¹.g; x/ W g 2 G; x 2 Aº. We call g the initial point and gx is the terminal point. That

212

Chapter 14 Free Groups and Group Presentations

is two points g; g1 in the vertex set are connected by an edge if g1 D gx for some x 2 A. We have .g; x/1 D .gx; x 1 /. This gives a directed graph called the Cayley graph of G on the generating set X. Call x the label on the edge .g; x/. Given a g 2 G then G is represented by at least one word W in A. This represents a path in the Cayley graph. The length of the word W is the length of the path. This is equivalent to making each edge have length one. If we take the distance between 2 points as the minimum path length we make the Cayley graph a metric space. This metric is called the word metric. If we extend this metric to all pairs of points in the Cayley graph in the obvious way (making each edge a unit real interval) then the Cayley graph becomes a geodesic metric space. Each closed path in the Cayley graph represents a relator. By left multiplication the group G acts on the Cayley graph as a group of isometries. Further the action of G on the Cayley graph is without inversion, that is ge ¤ e 1 , if e is an edge. If we sew in a 2-cell for each closed path in the Cayley graph we get a simply connected 2-complex called the Cayley complex.

14.6

Presentations of Factor Groups

Let G be a group with a presentation G D hXI Ri. Suppose that H is a factor group of G, that is H Š G=N for some normal subgroup N of G. We show that a presentation for H is then H D hX I R [ R1 i where R1 is a, perhaps additional, system of relators. Theorem 14.6.1 (Dyck’s theorem). Let G D hX I Ri and suppose that H Š G=N where N is a normal subgroup of G. Then a presentation for H is hXI R [ R1 i for some set of words R1 on X. Conversely the presentation hXI R [ R1 i defines a group that is a factor group of G. Proof. Since each element of H is a coset of N they have the form gN for g 2 G. It is clear then that the images of X generate H . Further since H is a homomorphic image of G each relator in R is a relator in H . Let N1 be a set of elements that generate N and let R1 be the corresponding words in the free group on X. Then R1 is an additional set of relators in H . Hence R [ R1 is a set of relators for H . Any relator in H is either a relator in G and hence a consequence of R or can be realized as an element of G that lies in N and hence is a consequence of R1 . Therefore R [ R1 is a complete set of defining relators for H and H has the presentation H D hX I R [R1 i. Conversely G D hXI Ri; G1 D hXI R [ R1 i. Then G D F .X /=N1 where N1 D N.R/ and G1 D F .X /=N2 where N2 D N.R [ R1 /. Hence N1  N2 . The normal subgroup N2 =N1 of F .X /=N1 corresponds to a normal subgroup of H of G and

Section 14.7 Group Presentations and Decision Problems

213

therefore by the isomorphism theorem G=H Š .F .X /=N1 /=.N2 =N1 / Š F .X /=N2 Š G1 :

14.7

Group Presentations and Decision Problems

We have seen that given any group G there exists a presentation for it, G D hXI Ri. In the other direction given any presentation hXI Ri we have seen that there is a group with that presentation. In principle every question about a group can be answered via a presentation. However things are not that simple. Max Dehn in his pioneering work on combinatorial group theory about 1910 introduced the following three fundamental group decision problems. (1) Word Problem: Suppose G is a group given by a finite presentation. Does there exist an algorithm to determine if an arbitrary word w in the generators of G defines the identity element of G? (2) Conjugacy Problem: Suppose G is a group given by a finite presentation. Does there exist an algorithm to determine if an arbitrary pair of words u; v in the generators of G define conjugate elements of G? (3) Isomorphism Problem: Does there exist an algorithm to determine given two arbitrary finite presentations whether the groups they present are isomorphic or not? All three of these problems have negative answers in general. That is for each of these problems one can find a finite presentation for which these questions cannot be answered algorithmically (see [23]). Attempts for solutions and for solutions in restricted cases have been of central importance in combinatorial group theory. For this reason combinatorial group theory has always searched for and studied classes of groups in which these decision problems are solvable. For finitely generated free groups there are simple and elegant solutions to all three problems. If F is a free group on x1 ; : : : ; xn and W is a freely reduced word in x1 ; : : : ; xn then W ¤ 1 if and only if L.W /  1. Since freely reducing any word to a freely reduced word is algorithmic this provides a solution to the word problem. Further a freely reduced word W D xve11 xve22 : : : xvenn is cyclically reduced if v1 ¤ vn or if v1 D vn then e1 ¤ en . Clearly then every element of a free group is conjugate to an element given by a cyclically reduced word called a cyclic reduction. This leads to a solution to the conjugacy problem. Suppose V and W are two words in the generators of F and V ; W are respective cyclic reductions. Then V is conjugate to W if and only if V is a cyclic permutation of W . Finally two finitely generated free groups are isomorphic if and only if they have the same rank.

214

14.8

Chapter 14 Free Groups and Group Presentations

Group Amalgams: Free Products and Direct Products

Closely related to free groups in both form and properties are free products of groups. Let A D ha1 ; : : : I R1 ; : : :i and B D hb1 ; : : : I S1 ; : : :i be two groups. We consider A and B to be disjoint. Then: Definition 14.8.1. The free product of A and B denoted A B is the group G with the presentation ha1 ; : : : ; b1 ; : : : I R1 ; : : : ; S1 ; : : :i, that is the generators of G consist of the disjoint union of the generators of A and B with relators taken as the disjoint union of the relators Ri of A and Sj of B. A and B are called the factors of G. In an analogous manner the concept of a free product can be extended to an arbitrary collection of groups. Definition 14.8.2. If A˛ D hgens A˛ I rels A˛ i; ˛ 2 I, is a collection of groups, then their free product G D A˛ is the group whose generators consist of the disjoint union of the generators of the A˛ and whose relators are the disjoint union of the relators of the A˛ . Free products exist and are nontrivial. We have: Theorem 14.8.3. Let G D A B. Then the maps A ! G and B ! G are injections. The subgroup of G generated by the generators of A has the presentation hgenerators of AI relators of Ai, that is, is isomorphic to A. Similarly for B. Thus A and B can be considered as subgroups of G. In particular A B is nontrivial if A and B are. Free products share many properties with free groups. First of all there is a categorical formulation of free products. Specifically we have: Theorem 14.8.4. A group G is the free product of its subgroups A and B if A and B generate G and given homomorphisms f1 W A ! H; f2 W B ! H into a group H there exists a unique homomorphism f W G ! H extending f1 and f2 . Secondly each element of a free product has a normal form related to the reduced words of free groups. If G D A B then a reduced sequence or reduced word in G is a sequence g1 g2 : : : gn , n  0, with gi ¤ 1, each gi in either A or B and gi ; gi C1 not both in the same factor. Then: Theorem 14.8.5. Each element g 2 G D A B has a unique representation as a reduced sequence. The length n is unique and is called the syllable length. The case n D 0 is reserved for the identity.

Section 14.8 Group Amalgams: Free Products and Direct Products

215

A reduced word g1 : : : gn 2 G D A B is called cyclically reduced if either n  1 or n  2 and g1 and gn are from different factors. Certainly every element of G is conjugate to a cyclically reduced word. From this we obtain several important properties of free products which are analogous to properties in free groups. Theorem 14.8.6. An element of finite order in a free product is conjugate to an element of finite order in a factor. In particular a finite subgroup of a free product is entirely contained in a conjugate of a factor. Theorem 14.8.7. If two elements of a free product commute then they are both powers of a single element or are contained in a conjugate of an abelian subgroup of a factor. Finally a theorem of Kurosh extends the Nielsen–Schreier theorem to free products. Theorem 14.8.8 (Kurosh). A subgroup of a free product is also a free product. Explicitly if G D A B and H  G then H D F . A˛ / . Bˇ / where F is a free group and . A˛ / is a free product of conjugates of subgroups of A and . Bˇ / is a free product of conjugates of subgroups of B. We note that the rank of F as well as the number of the other factors can be computed. A complete discussion of these is in [24], [23] and [15]. If A and B are disjoint groups then we now have two types of products forming new groups out of them, the free product and the direct product. In both these products the original factors inject. In the free product there are no relations between elements of A and elements of B while in a direct product each element of A commutes with each element of B. If a 2 A and b 2 B a cross commutator is Œa; b D aba1 b 1 . The direct product is a factor group of the free product and the kernel is precisely the normal subgroup generated by all the cross commutators. Theorem 14.8.9. Suppose that A and B are disjoint groups. Then A  B D .A ? B/=H where H is the normal closure in A ? B of all the cross commutators. In particular a presentation for A  B is given by A  B D hgens A; gens BI rels A; rels B; Œa; b for all a 2 A; b 2 Bi:

216

14.9

Chapter 14 Free Groups and Group Presentations

Exercises

1. Let X 1 be a set disjoint from X but bijective to X. A word in X is a finite sequence of letters from the alphabet. That is a word has the form i

i



w D xi11 xi2 2 : : : xinin ; where xij 2 X and ij D ˙1. Let W .X / be the set of all words on X. If w1 ; w2 2 W .X / we say that w1 is equivalent to w2 , denoted w1  w2 , if w1 can be converted to w2 by a finite string of insertions and deletions of trivial words. Verify that this is an equivalence relation on W .X /. 2. In F .X / let N.X / be the subgroup generated by all squares in F .X / that is N.X / D h¹g2 W g 2 F .X /ºi: Show that N.X / is a normal subgroup and the factor group F .X /=N.X / is abelian where every nontrivial element has order 2. 3. Show that a free group F is torsion-free. 4. Let F be a free group and a; b 2 F . Show: If ak D b k , k ¤ 0, then a D b. 5. Let F D ha; bI i a free group with basis ¹a; bº. Let ci D ai bai , i 2 Z. Then G D hci ; i 2 Zi is free with basis ¹ci ji 2 Zº. 6. Show that hx; yI x 2 y 3 ; x 3 y 4 i Š hxI xi D ¹1º. 7. Let G D hv1 ; : : : ; vn I v12    vn2 i, n  1, and ˛ W G ! Z2 the epimorphism with ˛.vi / D 1 for all i. Let U be the kernel of ˛. Then U has a presentation 1 1 U D hx1 ; : : : ; xn1 ; y1 ; : : : ; yn1 I y1 x1    yn1 xn1 yn1 xn1    y11 x11 i. 8. Let M D hx; yI x 2 ; y 3 i Š PSL.2; Z/ be the modular group. Let M 0 be the commutator subgroup. Show that M 0 is a free group of rank 2 with a basis ¹Œx; y; Œx; y 2 º.

Chapter 15

Finite Galois Extensions

15.1

Galois Theory and the Solvability of Polynomial Equations

As we mentioned in Chapter 1, one of the origins of abstract algebra was the problem of trying to determine a formula for finding the solutions in terms of radicals of a fifth degree polynomial. It was proved first by Ruffini in 1800 and then by Abel that it is impossible to find a formula in terms of radicals for such a solution. Galois in 1820 extended this and showed that such a formula is impossible for any degree five or greater. In proving this he laid the groundwork for much of the development of modern abstract algebra especially field theory and finite group theory. One of the goals of this book has been to present a comprehensive treatment of Galois theory and a proof of the results mentioned above. At this point we have covered enough general algebra and group theory to discuss Galois extensions and general Galois theory. In modern terms, Galois theory is that branch of mathematics that deals with the interplay of the algebraic theory of fields, the theory of equations and finite group theory. This theory was introduced by Evariste Galois about 1830 in his study of the insolvability by radicals of quintic (degree 5) polynomials, a result proved somewhat earlier by Ruffini and independently by Abel. Galois was the first to see the close connection between field extensions and permutation groups. In doing so he initiated the study of finite groups. He was the first to use the term group, as an abstract concept, although his definition was really just for a closed set of permutations. The method Galois developed not only facilitated the proof of the insolvability of the quintic and higher powers but led to other applications and to a much larger theory as well. The main idea of Galois theory is to associate to certain special types of algebraic field extensions called Galois extensions a group called the Galois group. The properties of the field extension will be reflected in the properties of the group, which are somewhat easier to examine. Thus, for example, solvability by radicals can be translated into solvability of groups which was discussed in Chapter 12. Showing that for every degree five or greater there exists a field extension whose Galois group is not solvable proves that there cannot be a general formula for solvability by radicals. The tie-in to the theory of equations is as follows: If f .x/ D 0 is a polynomial equation over some field F , we can form the splitting field K. This is usually a Galois

218

Chapter 15 Finite Galois Extensions

extension, and therefore has a Galois group called the Galois group of the equation. As before, properties of this group will reflect properties of this equation.

15.2

Automorphism Groups of Field Extensions

In order to define the Galois group we must first consider the automorphism group of a field extension. In this section K; L; M will always be (commutative) fields with additive identity 0 and multiplicative identity 1. Definition 15.2.1. Let LjK be a field extension. Then the set Aut.LjK/ D ¹˛ 2 Aut.L/ W ˛jK D the identity on Kº is called the set of automorphisms of L over K. Notice that if ˛ 2 Aut.LjK/ then ˛.k/ D k for all k 2 K. Lemma 15.2.2. Let LjK be a field extension. Then Aut.LjK/ forms a group called the Galois group of LjK. Proof. Aut.LjK/  Aut.L/ and hence to show that Aut.LjK/ is a group we only have to show that its a subgroup of Aut.L/. Now the identity map on L is certainly the identity map on K so 1 2 Aut.LjK/ and hence Aut.LjK/ is nonempty. If ˛; ˇ 2 Aut.LjK/ then consider ˛ 1 ˇ. If k 2 K then ˇ.k/ D k and ˛.k/ D k so ˛ 1 .k/ D k. Therefore ˛ 1 ˇ.k/ D k for all k 2 K and hence ˛ 1 ˇ 2 Aut.LjK/. It follows that Aut.LjK/ is a subgroup of Aut.L/ and therefore a group. If f .x/ 2 KŒx n K and L is the splitting field of f .x/ over K then Aut.LjK/ is also called the Galois group of f .x/. Theorem 15.2.3. If P is the prime field of L then Aut.LjP / D Aut.L/. Proof. We must show that any automorphism of a prime field P is the identity. If ˛ 2 Aut.L/ then ˛.1/ D 1 and so ˛.n  1/ D n  1. Therefore in P , ˛ fixes all integer multiples of the identity. However every element of P can be written as a quotient m1 n  1 of integer multiples of the identity. Since ˛ is a field homomorphism and ˛ fixes both the top and the bottom it follows that ˛ will fix every element of this form and hence fix each element of P . For splitting fields the Galois group is a permutation group on the roots of the defining polynomial. Theorem 15.2.4. Let f .x/ 2 KŒx and L the splitting field of f .x/ over K. Suppose that f .x/ has roots ˛1 ; : : : ; ˛n 2 L.

Section 15.2 Automorphism Groups of Field Extensions

219

(a) Then each  2 Aut.LjK/ is a permutation on the roots. In particular Aut.LjK/ is isomorphic to a subgroup of Sn and uniquely determined by the zeros of f .x/. (b) If f .x/ is irreducible then Aut.LjK/ operates transitively on ¹˛1 ; : : : ; ˛n º. Hence for each i; j there is a  2 Aut.LjK/ such that .˛i / D ˛j . (c) If f .x/ D b.x˛1 /    .x˛n / with ˛1 ; : : : ; ˛n distinct and Aut.LjK/ operates transitively on ˛1 ; : : : ; ˛n then f .x/ is irreducible. Proof. For the proofs we use the results of Chapter 8. (a) Let  2 Aut.K/. Then from Theorem 8.1.5 we obtain that  permutes the roots ˛1 ; : : : ; ˛n . Hence j¹˛1 ;:::;˛n º 2 Sn . This map then defines a homomorphism  W Aut.LjK/ ! Sn

by  ./ D j¹˛1 ;:::;˛n º :

Further  is uniquely determined by the images .˛i /. It follows that  is a monomorphism. (b) If f .x/ is irreducible then Aut.LjK/ operates transitively on the set ¹˛1 ; : : : ; ˛n º again following from Theorem 8.1.5. (c) Suppose that f .x/ D b.x  ˛1 /    .x  ˛n / with ˛1 ; : : : ; ˛n distinct and f 2 Aut.LjK/ operates transitively on ˛1 ; : : : ; ˛n . Assume that f .x/ D g.x/h.x/ with g.x/; h.x/ 2 KŒx n K. Without loss of generality let ˛1 be a zero of g.x/ and ˛n be a zero of h.x/. Let ˛ 2 Aut.LjK/ with ˛.˛1 / D ˛n . However ˛.g.x// D g.x/, that is ˛.˛1 / is a zero of ˛.g.x// D g.x/ which gives a contradiction since ˛n is not a zero of g.x/. Therefore f .x/ must be irreducible. p p Example 15.2.5. Let f .x/ D .x 2  2/.x 2  3/ 2 QŒx. The field L D Q. 2; 3/ is the spitting field of f .x/. Over L we have p p p p f .x/ D .x C 2/.x  2/.x C 3/.x  3/: We want to determine the Galois group Aut.LjQ/ D Aut.L/ D G. Lemma 15.2.6. The Galois group G above is the Klein 4-group. Proof. First we show  4. Let ˛ 2 Aut.L/. Then ˛ is uniquely p that jAut.L/j p determined by ˛. 2/ and ˛. 3/ and p 2 p p ˛.2/ D 2 D . 2/2 D ˛. 2 / D .˛. 2//2 : p p p p Hence ˛. 2/ D ˙ 2. Analogously ˛. 3/ D ˙ 3. From this it follows that jAut.L/j  4. Further ˛ 2 D 1 for any ˛ 2 G. p 3 is irreducible over K D Q. Next we show that the polynomial f .x/ D x 2  p p 2/. Assume that x 2  3 were reducible over K. Then 3 2 K. This implies that 3 D

220

Chapter 15 Finite Galois Extensions

p p C dc 2 with a; b; c; d 2 Z and b ¤ 0 ¤ d and gcd.c; d / D 1. Then bd 3 D p p c 2 C 2 2adbc. Since bd ¤ 0 this implies ad C bc 2 hence 3b 2 d 2 D a2 b 2 C 2b 2p that we must have ac D 0. If c D 0 then 3 D ab 2 Q a contradiction. If a D 0 then p p 3 D dc 2 which implies 3d 2 D 2c 2 . It follows from this that 3j gcd.c; d / D 1 p again a contradiction. Hence f .x/ D x 2  3 is irreducible over K D Q. 2/. Since L is the splitting field of f .x/ and p f .x/ is pirreducible over K then therepexists an automorphism ˛ 2 Aut.L/ with ˛. 3/ D  3 and ˛jK D IK , that is ˛. 2/ D p p p p p 2. Analogously there is a ˇ 2 Aut.L/ with ˇ. 2/ D  2 and ˇ. 3/ D 3. Clearly ˛ ¤ ˇ, ˛ˇ D ˇ˛ and ˛ ¤ ˛ˇ ¤ ˇ. It follows that Aut.L/ D ¹1; ˛; ˇ; ˛ˇº. completing the proof. a b

15.3

Finite Galois Extensions

We now define (finite) Galois extensions. First we introduce the concept of a Fix field. Let K be a field and G a subgroup of Aut.K/. Define the set Fix.K; G/ D ¹k 2 K W g.k/ D k 8g 2 Gº: Theorem 15.3.1. For a G  Aut.K/, the set Fix.K; G/ is a subfield of K called the Fix field of G over K. Proof. 1 2 K is in Fix.K; G/ so Fix.K; G/ is not empty. Let k1 ; k2 2 Fix.K; G/ and let g 2 G. Then g.k1 ˙ k2 / D g.k1 / ˙ g.k2 / since g is an automorphism. Then g.k1 / ˙ g.k2 / D k1 ˙ k2 and it follows that k1 ˙ k2 2 Fix.K; G/. In an analogous manner k1 k21 2 Fix.K; G/ if k2 ¤ 0 and therefore Fix.K; G/ is a subfield of K. Using the concept of a fix field we define a finite Galois extension. Definition 15.3.2. LjK is a (finite) Galois extension if there exists a finite subgroup G  Aut.L/ such that K D Fix.L; G/. We now give some examples of finite Galois extensions. p p Lemma 15.3.3. Let L D Q. 2; 3/ and K D Q. Then LjK is a Galois extension. Proof. Let G D Aut.LjK/. From the example in the previous section there are automorphisms ˛; ˇ 2 G with p p p p p p p p ˛. 3/ D  3; ˛. 2/ D 2 and ˇ. 2/ D  2; ˇ. 3/ D 3: We have

p p p p Q. 2; 3/ D ¹c C d 3 W c; d 2 Q. 2/º: p p p Let t D a1 C b1 2 C .a2 C b2 2/ 3 2 Fix.L; G/.

221

Section 15.4 The Fundamental Theorem of Galois Theory

Then applying ˇ we have

p p p t D ˇ.t / D a1  b1 2 C .a2  b2 2/ 3: p p It follows thatp b1 C b2 3 D 0, that is, b1 D b2 D 0 since p 3 … Q. Therefore t D a1 C a2 3. Applying ˛ we have ˛.t / D a1  a2 3 and hence a2 D 0. Therefore t D a1 2 Q. Hence Q D Fix.L; G/ and LjK is a Galois extension. 1

Lemma 15.3.4. Let L D Q.2 4 / and K D Q. Then LjK is not a Galois extension. 1

Proof. Suppose that ˛ 2 Aut.L/ and a D 2 4 . Then a is a zero of x 4  2 and hence 1

˛.a/ D 2 4

or 1

˛.a/ D i 2 4 … L ˛.a/ D 2

1 4

since i … L

or

or 1 4

˛.a/ D i 2 … L p p In particular ˛. 2/ D 2 and therefore

since i … L:

p Fix.L; Aut.L// D Q. 2/ ¤ Q:

15.4

The Fundamental Theorem of Galois Theory

We now state the fundamental theorem of Galois theory. This theorem describes the interplay between the Galois group and Galois extensions. In particular the result ties together subgroups of the Galois group and intermediate fields between L and K. Theorem 15.4.1 (fundamental theorem of Galois theory). Let LjK be a Galois extension with Galois group G D Aut.LjK/. For each intermediate field E let  .E/ be the subgroup of G fixing E. Then: (1)  is a bijection between intermediate fields containing K and subgroups of G. (2) LjK is a finite extension and if M is an intermediate field then jL W M j D jAut.LjM /j jM W Kj D jAut.LjK/ W Aut.LjM /j: (3) If M is an intermediate field then (a) LjM is always a Galois extension (b) M jK is a Galois extension if and only if Aut.LjM / is a normal subgroup of Aut.LjK/:

222

Chapter 15 Finite Galois Extensions

(4) If M is an intermediate field and M jK is a Galois extension then (a) ˛.M / D M for all ˛ 2 Aut.LjK/, (b) the map  W Aut.LjK/ ! Aut.M jK/ with .˛/ D ˛jM D ˇ is an epimorphism, (c) Aut.M jK/ D Aut.LjK/= Aut.LjM /. (5) The lattice of subfields of L containing K is the inverted lattice of subgroups of Aut.KjL/. We will prove this main result via a series of theorems and then combine them all. Theorem 15.4.2. Let G be a group, K a field and ˛1 ; : : : ; ˛n pairwise distinct group homomorphisms from G to K ? the multiplicative group of K. Then ˛1 ; : : : ; ˛n are linearly independent elements of the K-vector space of all homomorphisms from G to K. Proof. The proof is by induction on n. If n D 1 and k˛1 D 0 with k 2 K then 0 D k˛1 .1/ D k  1 and hence k D 0. Now suppose that n  2 and suppose that each n  1 of the ˛1 ; : : : ; ˛n are linearly independent over K. If n X

ki ˛i D 0;

ki 2 K

( )

i D1

then we must show that all ki D 0. Since ˛1 ¤ ˛n there exists an a 2 G with ˛1 .a/ ¤ ˛n .a/. Let g 2 G and apply the sum above to ag. We get n X

ki .˛i .a//.˛i .g// D 0:

(

)

i D1

Now multiply equation ( ) by ˛n .a/ 2 K to get n X

ki .˛n .a//.˛i .g// D 0:

(

)

i D1

If we subtract equation (

) from equation (

) then the last term vanishes and we have an equation in the n  1 homomorphism ˛1 ; : : : ; ˛n1 . Since these are linearly independent we obtain k1 .˛1 .a//  k1 .˛n .a// D 0 for the coefficient for ˛1 . Since ˛1 .a/ ¤ ˛n .a/ we must have k1 D 0. Now ˛2 ; : : : ; ˛n1 are by assumption linearly independent so k2 D    D kn D 0 also. Hence all the coefficients must be zero and therefore the mappings are independent.

223

Section 15.4 The Fundamental Theorem of Galois Theory

Theorem 15.4.3. Let ˛1 ; : : : ; ˛n be pairwise distinct monomorphisms from the field K into the field K 0 . Let L D ¹k 2 K W ˛1 .k/ D ˛2 .k/ D    D ˛n .k/º: Then L is a subfield of K with jL W Kj  n. Proof. Certainly L is a field. Assume that r D jK W Lj < n and let ¹a1 ; : : : ; ar º be a basis of the L-vector space K. We consider the following system of linear equations with r equations and n unknowns. .˛1 .a1 //x1 C    C .˛n .a1 //xn D 0 :: : .˛1 .ar //x1 C    C .˛n .ar //xn D 0: Since r < n there exists a nontrivial solution .x1 ; : : : ; xn / 2 .K 0 /n . Let a 2 K. Then r X lj aj with lj 2 L: aD j D1

From the definition of L we have ˛1 .lj / D ˛i .lj / for i D 2; : : : ; n: Then with our nontrivial solution .x1 ; : : : ; xn / we have n X

xi .˛i .a// D

i D1

n X

xi

i D1

D

r X

X r

 ˛i .lj /˛i .aj /

j D1

.˛1 .lj //

j D1

n X

xi .˛i .aj // D 0

i D1

since Pn ˛1 .lj / D ˛i .lj / for i D 2; : : : ; n. This holds for all a 2 K and hence iD1 xi ˛i D 0 contradicting Theorem 15.4.2. Therefore our assumption that jK W Lj < n must be false and hence jK W Lj  n. Definition 15.4.4. Let K be a field and G a finite subgroup of Aut.K/. The map trG W K ! K given by X ˛.k/ trG .k/ D ˛2G

is called the G-trace of K.

224

Chapter 15 Finite Galois Extensions

Theorem 15.4.5. Let K be a field and G a finite subgroup of Aut.K/. Then ¹0º ¤ trG .K/  Fix.K; G/: Proof. Let ˇ 2 G. Then ˇ.trG .k// D

X ˛2G

ˇ˛.k/ D

X

˛.k/ D trG .k/:

˛2G

Therefore trG .K/  Fix.K; G/. P Now assumePthat trG .k/ D 0 for all k 2 K. Then ˛2G ˛.k/ D 0 for all k 2 K. It follows that ˛2G ˛ is the zero map and hence the set of all ˛ 2 G are linearly dependent as elements of the K-vector space of all maps from K to K. This contradicts Theorem 15.4.2 and hence the trace cannot be the zero map. Theorem 15.4.6. Let K be a field and G a finite subgroup of Aut.K/. Then jK W Fix.K; G/j D jGj: Proof. Let L D Fix.K; G/ and suppose that jGj D n. From Theorem 15.4.3 we know that jK W Lj  n. We must show that jK W Lj  n. Suppose that G D ¹˛1 ; : : : ; ˛n º. To prove the result we show that if m > n and a1 ; : : : ; am 2 K then a1 ; : : : ; am are linearly dependent. We consider the system of equations .˛11 .a1 //x1 C    C .˛11 .am //xm D 0 :: : .˛n1 .a1 //x1 C    C .˛n1 .am //xm D 0: Since m > n there exists a nontrivial solution .y1 ; : : : ; ym / 2 K m . Suppose that yl ¤ 0. Using Theorem 15.4.5 we can choose k 2 K with trG .k/ ¤ 0. Define .x1 ; : : : ; xm / D kyl1 .y1 ; : : : ; ym /: This m-tuple .x1 ; : : : ; xm / is then also a nontrivial solution of the system of equations considered above. Then we have trG .xl / D trG .k/ since xl D k: Now we apply ˛i to the i-th equation to obtain a1 .˛1 .x1 // C    C am .˛1 .xm // D 0 :: : a1 .˛n .x1 // C    C am .˛n .xm // D 0:

Section 15.4 The Fundamental Theorem of Galois Theory

225

Summation leads to 0D

m X j D1

aj

n m X X .˛i .xj // D .trG .xj //aj i D1

j D1

by definition of the G-trace. Hence a1 ; : : : ; am are linearly dependent over L since trG .xl / ¤ 0. Therefore jK W Lj  n. Combining this with Theorem 15.4.3 we get that jK W Lj D n D jGj. Theorem 15.4.7. Let K be a field and G a finite subgroup of Aut.K/. Then Aut.Kj Fix.K; G// D G: Proof. G  Aut.Kj Fix.K; G// since if g 2 G then g 2 Aut.K/ and g fixes Fix.K; G/ by definition. Therefore we must show that Aut.Kj Fix.K; G//  G. Assume then that there exists an ˛ 2 Aut.Kj Fix.K; G// with ˛ … G. Suppose, as in the previous proof, jGj D n and G D ¹˛1 ; : : : ; ˛n º with ˛1 D 1. Now Fix.K; G/ D ¹a 2 K W a D ˛2 .a/ D    D ˛n .a/º D ¹a 2 K W ˛.a/ D a D ˛2 .a/ D    D ˛n .a/º: From Theorem 15.4.3 we have that jK W Fix.K; G/j  n C 1. However from Theorem 15.4.6 jK W Fix.K; G/j D n getting a contradiction. Suppose that LjK is a Galois extension. We now establish that the map  between intermediate fields K  E  L and subgroups of Aut.LjK/ is a bijection. Theorem 15.4.8. Let LjK be a Galois extension. Then: (1) Aut.LjK/ is finite and Fix.L; Aut.LjK// D K: (2) If H  Aut.LjK/ then Aut.Lj Fix.L; H // D H: Proof. (1) If .LjK/ is a Galois extension there is a finite subgroup of Aut.L/ with K D Fix.K; G/. From Theorem 15.4.7 we have G D Aut.LjK/. In particular Aut.LjK/ is finite and K D Fix.L; Aut.LjK//. (2) Let H  Aut.LjK/. From part (1) H is finite and then Aut.Lj Fix.L; H // D H from Theorem 15.4.7.

226

Chapter 15 Finite Galois Extensions

Theorem 15.4.9. Let LjK be a field extension. Then the following are equivalent. (1) LjK is a Galois extension. (2) jL W Kj D jAut.LjK/j < 1. (3) jAut.LjK/j < 1 and K D Fix.L; Aut.LjK//. Proof. (1) ) (2): Now jAut.LjK/j < 1 and Fix.L; Aut.LjK// D K from Theorem 15.4.8. Therefore jL W Kj D jAut.LjK/j from Theorem 15.4.6. (2) ) (3): Let G D Aut.LjK/. Then K  Fix.L; G/  L. From Theorem 15.4.6 we have jL W Fix.L; G/j D jGj D jL W Kj: (3) ) (1) follows directly from the definition completing the proof. We now show that if LjK is a Galois extension then LjM is also a Galois extension for any intermediate field M . Theorem 15.4.10. Let LjK be a Galois extension and K  M  L be an intermediate field. Then LjM is always a Galois extension and jM W Kj D jAut.LjK/ W Aut.LjM /j: Proof. Let G D Aut.LjK/. Then jGj < 1 and further K D Fix.L; G/ from Theorem 15.4.9. Define H D Aut.LjM / and M 0 D Fix.L; H /. We must show that M 0 D M for then LjM is a Galois extension. S Since the elements of H fix M we have M  M 0 . Let G D riD1 ˛i H a disjoint union of the cosets of H . Let ˛1 D 1 and define ˇi D .˛i /jM . The ˇ1 ; : : : ; ˇr are pairwise distinct for if ˇi D ˇj , that is .˛i /jM D .˛j /jM , then ˛j1 ˛i 2 H so ˛i and ˛j are in the same coset. We claim that ¹a 2 M W ˇ1 .a/ D    D ˇr .a/º D M \ Fix.L; G/: Further we know that M \ Fix.L; G/ D M \ K D K from Theorem 15.4.9. To establish the claim it is clear that M \ Fix.L; G/  ¹a 2 M W ˇ1 .a/ D    D ˇr .a/º since a D ˇi .a/ D ˛i .a/

for ˛i 2 G; a 2 K:

Section 15.4 The Fundamental Theorem of Galois Theory

227

Hence we must show that ¹a 2 M W ˇ1 .a/ D    D ˇr .a/º  M \ Fix.L; G/: To do this we must show that ˛.b/ D b for all ˛ 2 G, b 2 M . We have ˛ 2 ˛i H for some i and hence ˛ D ˛i for 2 H . We obtain then ˛.b/ D ˛i . .b// D ˛i .b/ D ˇi .b/ D b proving the inclusion and establishing the claim. Now jM W Kj  r from Theorem 15.4.3. From the degree formula we get jL W M 0 jjM 0 W M jjM W Kj D jL W Kj D jGj D jG W H jjH j D rjL W M 0 j since jL W Kj D jGj from Theorem 15.4.9 and jH j D jL W M 0 j from Theorem 15.4.6. Therefore jM W M 0 j D 1 and hence M D M 0 since jM W Kj  r. Now jM W Kj D jG W H j D jAut.LjK/ W Aut.LjM /j completing the proof. Lemma 15.4.11. Let LjK be a field extension and K  M  L be an intermediate field. If ˛ 2 Aut.LjK/ then Aut.Lj˛.M // D ˛ Aut.LjM /˛ 1 : Proof. Now ˇ 2 Aut.Lj˛.M // if and only if ˇ.˛.a// D ˛.a/ for all a 2 M . This occurs if and only if ˛ 1 ˇ˛.a/ D a for all a 2 M which is true if and only if ˇ 2 ˛ Aut.LjM /˛ 1 . Lemma 15.4.12. Let LjK be a Galois extension and K  M  L be an intermediate field. Suppose that ˛.M / D M for all ˛ 2 Aut.LjK/. Then  W Aut.LjK/ ! Aut.M jK/

by .˛/ D ˛jM

is an epimorphism with kernel ker./ D Aut.LjM /. Proof. It is clear that  is a homomorphism with ker./ D Aut.LjM / (see exercises). We must show that it is an epimorphism. Let G D im./. Since LjK is a Galois extension we get that Fix.M; G/ D Fix.L; Aut.LjK// \ M D K \ M D K: Then from Theorem 15.4.8 we have Aut.M jK/ D Aut.M j Fix..M; G// D G and therefore  is an epimorphism.

228

Chapter 15 Finite Galois Extensions

Theorem 15.4.13. Let LjK be a Galois extension and K  M  L be an intermediate field. Then the following are equivalent. (1) M jK is a Galois extension. (2) If ˛ 2 Aut.LjK/ then ˛.M / D M . (3) Aut.LjM / is a normal subgroup of Aut.LjK/. Proof. (1) ) (2): Suppose that M jK is a Galois extension. Let Aut.M jK/ D ¹˛1 ; : : : ; ˛r º. Consider the ˛i as monomorphisms from M into L. Let ˛rC1 W M ! L be a monomorphism with ˛rC1jK D 1. Then ¹a 2 M W ˛1 .a/ D ˛2 .a/    D ˛r .a/ D ˛rC1 .a/º D K since M jK is a Galois extension. Therefore from Theorem 15.4.3 we have that, if the ˛1 ; : : : ; ˛r ; ˛rC1 are distinct then jM W Kj  r C 1 > r D jAut.M jK/j D jM W Kj giving a contradiction. Hence if ˛rC1 2 Aut.LjK/ is arbitrary then ˛rC1jM 2 ¹˛1 ; : : : ; ˛r º, that is ˛rC1 fixes M . (2) ) (1): Suppose that if ˛ 2 Aut.LjK/ then ˛.M / D M . The map  W Aut.LjK/ ! Aut.M jK/ with .˛/ D ˛jM is surjective. Since LjK is a Galois extension then Aut.LjK/ is finite. Therefore also H D Aut.M jK/ is finite. To prove (1) then it is sufficient to show that K D Fix.M; H /. The field K  Fix.M; H / from the definition of the Fix field. Hence we must show that Fix.M; H /  K. Assume that there exists an ˛ 2 Aut.LjK/ with ˛.a/ ¤ a. Recall that LjK is a Galois extension and therefore Fix.L; Aut.LjK// D K. Define ˇ D ˛jM . Then ˇ 2 H since ˛.M / D M and our original assumption. Then ˇ.a/ ¤ a contradicting a 2 Fix.M; H /. Therefore K D Fix.M; H / and M jK is a Galois extension. (2) ) (3): Suppose that if ˛ 2 Aut.LjK/ then ˛.M / D M . Then Aut.LjM / is a normal subgroup of Aut.LjK/ follows from Lemma 15.4.12 since Aut.LjM / is the kernel of . (3) ) (2): Suppose that Aut.LjM / is a normal subgroup of Aut.LjK/. Let ˛ 2 Aut.LjK/ then from our assumption and Lemma 15.4.11 we get that Aut.Lj˛.M // D Aut.LjM /: Now LjM and Lj˛.M / are Galois extensions by Theorem 15.4.10. Therefore ˛.M / D Fix.L; Aut.Lj˛.M // D Fix.L; Aut.LjM // D M completing the proof.

Section 15.4 The Fundamental Theorem of Galois Theory

229

We now combine all of these results to give the proof of Theorem 15.4.1, the fundamental theorem of Galois theory. Proof of Theorem 15.4.1. Let LjK be a Galois extension. (1) Let G  Aut.LjK/. Both G and Aut.LjK/ are finite from Theorem 15.4.8. Further G D Aut.Lj Fix.L; G// from Theorem 15.4.7. Now let M be an intermediate field of LjK. Then LjM is a Galois extension from Theorem 15.4.10 and then Fix.L; Aut.LjM // D M from Theorem 15.4.8. (2) Let M be an intermediate field of LjK. From Theorem 15.4.10 LjM is a Galois extension. From Theorem 15.4.9 we have jL W M j D jAut.LjM /j. Applying Theorem 15.4.10 we get the result on indices jM W Kj D jAut.LjK/ W Aut.LjM /j: (3) Let M be an intermediate field of LjK. (a) From Theorem 15.4.10 we have that LjM is a Galois extension. (b) From Theorem 15.4.13 M jK is a Galois extension if and only if Aut.LjM / is a normal subgroup of Aut.LjK/: (4) Let M jK be a Galois extension. (a) ˛.M / D M for all ˛ 2 Aut.LjK/ from Theorem 15.4.13. (b) The map  W Aut.LjK/ ! Aut.M jK/ with .˛/ D ˛jM D ˇ is an epimorphism follows from Lemma 15.4.12 and Theorem 15.4.13. (c) Aut.M jK/ D Aut.LjK/= Aut.LjM / follows directly from the group isomorphism theorem. (5) That the lattice of subfields of L containing K is the inverted lattice of subgroups of Aut.LjK/ follows directly from the previous results. In Chapter 8 we looked at the following example (Example 8.1.7). Here we analyze it further using the Galois theory. Example 15.4.14. Let f .x/ D x 3  7 2 QŒx. This has no zeros in Q and since it is of degree 3 it follows that it must be irreducible in QŒx. p 3 1 Let ! D  2 C 2 i 2 C. Then it is easy to show by computation that p 3 1 i ! D  2 2 2

and

! 3 D 1:

Therefore the three zeros of f .x/ in C are a1 D 71=3 ;

a2 D !.71=3 /;

a3 D ! 2 .71=3 /:

230

Chapter 15 Finite Galois Extensions

Hence L D Q.a1 ; a2 ; a3 / is the splitting field of f .x/. Since the minimal polynomial of all three zeros over Q is the same .f .x// it follows that Q.a1 / Š Q.a2 / Š Q.a3 /: Since Q.a1 /  R and a2 ; a3 are nonreal it is clear that a2 ; a3 … Q.a1 /. Suppose that Q.a2 / D Q.a3 /. Then ! D a3 a21 2 Q.a2 / and so 71=3 D ! 1 a2 2 Q.a2 /. Hence Q.a1 /  Q.a2 / and therefore Q.a1 / D Q.a2 / since they are the same degree over Q. This contradiction shows that Q.a2 / and Q.a3 / are distinct. By computation we have a3 D a11 a22 and hence L D Q.a1 ; a2 ; a3 / D Q.a1 ; a2 / D Q.71=3 ; !/: Now the degree of L over Q is jL W Qj D jQ.71=3 ; !/ W Q.!/jjQ.!/ W Qj: Now jQ.!/ W Qj D 2 since the minimal polynomial of ! over Q is x 2 C x C 1. Since no zero of f .x/ lies in Q.!/ and the degree of f .x/ is 3 it follows that f .x/ is irreducible over Q.!/. Therefore we have that the degree of L over Q.!/ is 3. Hence jL W Qj D .2/.3/ D 6. Clearly then we have the following lattice of intermediate fields:

The question then arises as to whether these are all the intermediate fields. The answer is yes which we now prove. Let G D Aut.LjQ/ D Aut.L/. (Aut.LjQ/ D Aut.L/ since Q is a prime field). Now G Š S3 . G acts transitively on ¹a1 ; a2 ; a3 º since f is irreducible. Let ı W C ! C be the automorphism of C taking each element to its complex conjugate, that is ı.z/ D z. Then ı.f / D f and ıjL 2 G (see Theorem 8.2.2). Since a1 2 R we get that ıj¹a1 ;a2 ;a3 º D .a2 ; a3 / the 2-cycle that maps a2 to a3 and a3 to a2 . Since G is transitive on ¹a1 ; a2 ; a3 º there is a  2 G with  .a1 / D a2 .

Section 15.5 Exercises

231

Case 1:  .a3 / D a3 . Then  D .a1 ; a2 / and .a1 ; a2 /.a2 ; a3 / D .a1 ; a2 ; a3 / 2 G. Case 2:  .a3 / ¤ a3 . Then  is a 3-cycle. In either case G is generated by a transposition and a 3-cycle. Hence G is all of S3 . Then LjQ is a Galois extension from Theorem 15.4.9 since jGj D jL W Qj. The subgroups of S3 are as follows:

Hence the above lattice of fields is complete. LjQ; QjQ; Q.!/jQ and LjQ.ai / are Galois extensions while Q.ai /jQ with i D 1; 2; 3 are not Galois extensions.

15.5

Exercises

1. Let K  M  L be a chain of fields and let  W Aut.LjK/ ! Aut.M jK/ be defined by .˛/ D ˛jM . Show that  is an epimorphism with kernel ker./ D Aut.LjM /. p p 1 1 2. Show that Q.5 4 /jQ. 5/ and Q. 5/jQ are Galois extensions and Q.5 4 /jQ is not a Galois extension. 3. Let LjK be a field extension and u; v 2 L algebraic over K with jK.u/ W Kj D m and jK.v/ W Kj D n. If m and n are coprime, then jK.u; v/ W Kj D n  m. 1 p 4. Let p; q be prime numbers with p ¤ q. Let L D Q. p; q 3 /. Show that 1 p L D Q. p  q 3 /. Determine a basis of L over Q and the minimal polynomial 1 p of p  q 3 . 1

5. Let K D Q.2 n / with n  2. (i) Determine the number of Q-embeddings W K ! R. Show that for each such embedding we have .K/ D K. (ii) Determine Aut.KjQ/.

232

Chapter 15 Finite Galois Extensions

6. Let ˛ D

p p 5 C 2 5.

(i) Determine the minimal polynomial of ˛ over Q. (ii) Show that Q.a/jQ is a Galois extension. (iii) Determine Aut.Q.a/jQ/. 7. Let K be a field of prime characteristic p and let f .x/ D x p  x C a 2 K be an irreducible polynomial. Let L D K.v/, where v is a zero of f .x/. (i) If ˛ is a zero of f .x/ then also ˛ C 1. (ii) LjK is a Galois extension. (iii) There is exactly one K-automorphism of L with .v/ D v C 1. (iv) The Galois group Aut.LjK/ is cyclic with generating element .

Chapter 16

Separable Field Extensions

16.1

Separability of Fields and Polynomials

In the previous chapter we introduced and examined Galois extensions. Recall that LjK is a Galois extension if there exists a finite subgroup G  Aut.L/ such that K D Fix.L; G/. The following questions immediately arise. (1) Under what conditions is a field extension LjK a Galois extension? (2) When is LjK a Galois extension when L is the splitting field of a polynomial f .x/ 2 KŒx? In this chapter we consider these questions and completely characterize Galois extensions. In order to do this we must introduce separable extensions. Definition 16.1.1. Let K be a field. Then a nonconstant polynomial f .x/ 2 KŒx is called separable over K if each irreducible factor of f .x/ has only simple zeros in its splitting field. We now extend this definition to field extensions. Definition 16.1.2. Let LjK be a field extension and a 2 L. Then a is separable over K if a is a zero of a separable polynomial. The field extension LjK is a separable field extension or just separable if all a 2 L are separable over K. In particular a separable extension is an algebraic extension. Finally we consider fields where every nonconstant polynomial is separable. Definition 16.1.3. A field K is perfect if each nonconstant polynomial in KŒx is separable over K. The following is straightforward from the definitions. An element a is separable over K if and only if its minimal polynomial ma .x/ is separable. P If f .x/ 2 KŒx then P f .x/ D niD0 ki x i with ki 2 K. The formal derivative of f .x/ is then f 0 .x/ D niD1 i ki x i 1 . As in ordinary Calculus we have the usual differentiation rules .f .x/ C g.x//0 D f 0 .x/ C g 0 .x/ and

.f .x/g.x//0 D f 0 .x/g.x/ C f .x/g 0 .x/

for f .x/; g.x/ 2 KŒx.

234

Chapter 16 Separable Field Extensions

Lemma 16.1.4. Let K be a field and f .x/ an irreducible nonconstant polynomial in KŒx. Then f .x/ is separable if and only if its formal derivative is nonzero. Proof. Let L be the splitting field of f .x/ over K. Let f .x/ D .x  a/r g.x/ where .x  a/ does not divide g.x/. Then f 0 .x/ D .x  a/r1 .rg.x/ C .x  a/g 0 .x//: If f 0 .x/ ¤ 0 then a is a zero of f .x/ in L over K of multiplicity m  2 if and only if .x  a/jf .x/ and also .x  a/jf 0 .x/. Let f .x/ be a separable polynomial over KŒx and let a be a zero of f .x/ in L. Then if f .x/ D .x  a/r g.x/ with .x  a/ not dividing g.x/ we must have r D 1. Then f 0 .x/ D g.x/ C .x  a/g 0 .x/: If g 0 .x/ D 0 then f 0 .x/ D g.x/ ¤ 0. Now suppose that g0 .x/ ¤ 0. Assume that f 0 .x/ D 0. Then necessarily .x  a/jg.x/ giving a contradiction. Therefore f 0 .x/ ¤ 0. Conversely suppose that f 0 .x/ ¤ 0. Assume that f .x/ is not separable. Then both f .x/ and f 0 .x/ have a common zero a 2 L. Let ma .x/ be the minimal polynomial of a in KŒx. Then ma .x/jf .x/ and ma .x/jf 0 .x/. Since f .x/ is irreducible then the degree of ma .x/ must equal the degree of f .x/. But ma .x/ must also have the same degree as f 0 .x/ which is less than that of f .x/ giving a contradiction. Therefore f .x/ must be separable. We now consider the following example of a nonseparable polynomial over the finite field Zp of p elements. We will denote this field now as GF.p/, the Galois field of p elements. Example 16.1.5. Let K D GF.p/ and L D K.t / the field of rational functions in t over K. Consider the polynomial f .x/ D x p  t 2 LŒx. Now KŒt =tKŒt  Š K. Since K is a field this implies that tKŒt  is a maximal ideal and hence a prime ideal in KŒt  with prime element t 2 KŒt  (see Theorem 3.2.7). By the Eisenstein criteria f .x/ is an irreducible polynomial in LŒx (see Theorem 4.4.8). However f 0 .x/ D px p1 D 0 since char.K/ D p. Therefore f .x/ is not separable.

16.2

Perfect Fields

We now consider when a field K is perfect. First we show that in general any field of characteristic 0 is perfect. In particular the rationals Q are perfect and hence any extension of the rationals is separable. Theorem 16.2.1. Each field K of characteristic zero is perfect.

Section 16.2 Perfect Fields

235

Proof. Suppose that K is a field with char.K/ D 0. Suppose that f .x/ is a nonconstant polynomial in KŒx. Then f 0 .x/ ¤ 0. If f .x/ is irreducible then f .x/ is separable from Lemma 16.1.4. Therefore by definition each nonconstant polynomial f .x/ 2 KŒx is separable. We remark that in the original motivation for Galois theory the ground field was the rationals Q. Since this has characteristic zero it is perfect and all extensions are separable. Hence the question of separability didn’t arise until the question of extensions of fields of prime characteristic arose. Corollary 16.2.2. Any finite extension of the rationals Q is separable. We now consider the case of prime characteristic. Theorem 16.2.3. Let K be a field with char.K/ D p ¤ 0. If f .x/ is a nonconstant polynomial in KŒx then the following are equivalent: (1) f 0 .x/ D 0. (2) f .x/ is a polynomial in x p , that is, there is a g.x/ 2 KŒx with f .x/ D g.x p /. If in .1/ and .2/ f .x/ is irreducible then f .x/ is not separable over K if and only if f .x/ is a polynomial in x p . P Proof. Let f .x/ D niD1 ai x i . Then f 0 .x/ D 0 if and only if pji for all i with ai ¤ 0. But this is equivalent to f .x/ D a0 C ap x p C    C am x mp : If f .x/ is irreducible then we have that f .x/ is not separable if and only if f 0 .x/ D 0 from Lemma 16.1.4. Theorem 16.2.4. Let K be a field with char.K/ D p ¤ 0. Then the following are equivalent: (1) K is perfect. (2) Each element in K has a p-th root in K. (3) The Frobenius homomorphism x 7! x p is an automorphism of K. Proof. First we show that (1) implies (2). Suppose that K is perfect and a 2 K. Then x p  a is separable over K. Let g.x/ 2 KŒx be an irreducible factor of x p  a. Let L be the splitting field of g.x/ over K and b a zero of g.x/ in L. Then b p D a. Further x p b p D .x b/p 2 LŒx since the characteristic of K is p. Hence g.x/ D .x b/s and then s must equal 1 since g.x/ is irreducible. Therefore b 2 K and b is a p-th root of a. Now we show that (2) implies (3). Recall that the Frobenius homomorphism  W x 7! x p is injective (see Theorem 1.8.8). We must show that it is also surjective. Let

236

Chapter 16 Separable Field Extensions

a 2 K and let b be a p-th root of a so that a D b p . Then  .b/ D b p D a and  is surjective. Finally we show that (3) implies (1). Let  W x 7! x p be surjective. It follows that each a 2 K has a p-th root in K. Now let f .x/ 2 KŒx be irreducible. Assume that f .x/ is not separable. From Theorem 16.2.3 there is a g.x/ 2 KŒx with f .x/ D g.x p /, that is f .x/ D a0 C a1 x p C    C am x mp : Let bi 2 K with ai D bip . Then p p mp x D .b0 C b1 x C    C bm x m /p : f .x/ D bop C b1 x p C    C bm

However this is a contradiction since f .x/ is irreducible. Therefore f .x/ is separable completing the proof. Theorem 16.2.5. Let K be a field with char.K/ D p ¤ 0. Then each element of K has at most one p-th power in K. p p Proof. Suppose that b1 ; b2 2 K with b1 D b2 D a. Then

0 D b1p  b2p D .b1  b2 /p : Since K has no zero divisors it follows that b1 D b2 .

16.3

Finite Fields

In this section we consider finite fields. In particular we show that if K is a finite field then jKj D p m for some prime p and natural number m > 0. Further we show that if K1 ; K2 are finite fields with jK1 j D jK2 j then K1 Š K2 . Hence there is a unique finite field for each possible order. Notice that if K is a finite field then by necessity char K D p ¤ 0. We first show that in this case K is always perfect. Theorem 16.3.1. A finite field is perfect. Proof. Let K be a finite field of characteristic p > 0. Then the Frobenius map  W x 7! x p is surjective since its injective and K is finite. Therefore K is perfect from Theorem 16.2.4. Next we show that each finite field has order p m for some prime p and natural number m > 0. Lemma 16.3.2. Let K be a finite field. Then jKj D p m for some prime p and natural number m > 0.

237

Section 16.3 Finite Fields

Proof. Let K be a finite field with characteristic p > 0. Then K can be considered as a vector space over K D GF.p/ and hence of finite dimension since jKj < 1. If ˛1 ; : : : ; ˛m is a basis then each f 2 K can be written as f D c1 ˛1 C    C cn ˛m with each ci 2 GF.p/. Hence there are p choices for each ci and therefore p m choices for each f . In Theorem 9.5.16 we proved that any finite subgroup of the multiplicative group of a field is cyclic. If K is a finite field then its multiplicative subgroup K ? is finite and hence is cyclic. Lemma 16.3.3. Let K be a finite field. Then its multiplicative subgroup K ? is cyclic. If K is a finite field with order p m then its multiplicative subgroup K ? has order p  1. Then from Lagrange’s theorem each nonzero element to the power p m is the identity. Therefore we have the result. m

Lemma 16.3.4. Let K be a field of order p m . Then each ˛ 2 K is a zero of the m m polynomial x p  x. In particular if ˛ ¤ 0 then ˛ is a zero of x p 1  1. If K is a finite field of order p m , it is a finite extension of GF.p/. Since the multiplicative group is cyclic we must have K D GF.p/.˛/ for some ˛ 2 K. From this we obtain that for a given possible finite order there is only one finite field up to isomorphism. Theorem 16.3.5. Let K1 ; K2 be finite fields with jK1 j D jK2 j. Then K1 Š K2 . Proof. Let jK1 j D jK2 j D p m . From the remarks above K1 D GF.p/.˛/ where ˛ has order p m  1 in K1? . Similarly K2 D GF.p/.ˇ/ where ˇ also has order p m  1 in K2? . Hence GF.p/.˛/ Š GF.p/.ˇ/ and therefore K1 Š K2 . In Lemma 16.3.2 we saw that if K is a finite field then jKj D pn for some prime p and positive integer n. We now show that given a prime power p n there does exist a finite field of that order. Theorem 16.3.6. Let p be a prime and n > 0 a natural number. Then there exists a field K of order p n . n

Proof. Given a prime p consider the polynomial g.x/ D x p  x 2 GF.p/Œx. Let K be the splitting field of this polynomial over GF.p/. Since a finite field is perfect K is a separable extension and hence all the zeros of g.x/ are distinct in K. Let F be the set of p n distinct zeros of g.x/ within K. Let a; b 2 F . Since n

n

.a ˙ b/p D ap ˙ b p

n

and

n

n

.ab/p D ap b p

n

238

Chapter 16 Separable Field Extensions

it follows that F forms a subfield of K. However F contains all the zeros of g.x/ and since K is the smallest extension of GF.p/ containing all the zeros of g.x/ we must have K D F . Since F has p n elements it follows that the order of K is p n . Combining Theorems 16.3.5 and 16.3.6 we get the following summary result indicating that up to isomorphism there exists one and only one finite field of order p n . Theorem 16.3.7. Let p be a prime and n > 0 a natural number. Then up to isomorphism there exists a unique finite field of order p n .

16.4

Separable Extensions

In this section we consider some properties of separable extensions. Theorem 16.4.1. Let K be a field with K  L and L algebraically closed. Let ˛ W K ! L be a monomorphism. Then the number of monomorphisms ˇ W K.a/ ! L with ˇjK D ˛ is equal to the number of pairwise distinct zeros in L of the minimal polynomial ma of a over K. Proof. Let ˇ be as in the statement of the theorem. Then ˇ is uniquely determined by ˇ.a/ and ˇ.a/ is a zero of the polynomial ˇ.ma .x// D ˛.ma .x//. Now let a0 be a zero of ˛.ma .x// in L. Then there exists a ˇ W K.a/ ! L with ˇ.a/ D a0 from Theorem 7.1.4. Therefore ˛ has exactly as many extensions ˇ as ˛.ma .x// has pairwise distinct zeros in L. The number of pairwise distinct zeros of ˛.ma .x// is equal to the number of pairwise distinct zeros of ma .x/. This can be seen as follows. Let L0 be a splitting field of ma .x/ and L1  L a splitting field of ˛.ma .x//. From Theorems 8.1.5 and 8.1.6 there is an isomorphism W L0 ! L1 which maps the zeros of ma .x/ onto the zeros of ˛.ma .x//. Lemma 16.4.2. Let LjK be a finite extension with L  L and L algebraically closed. In particular L D K.a1 ; : : : ; an / where the ai are algebraic over K. Let ki be the number of pairwise distinct zeros of the minimal polynomial mai of ai over K.a1 ; : : : ; an1 / in L. Then there are exactly k1 ; : : : ; kn monomorphisms ˇ W L ! L with ˇjK D 1K . Proof. From Theorem 16.4.1 there are exactly p1 monomorphisms ˛ W K.a1 / ! L with ˛jK equal to the identity on K. Each such ˛ has exactly p2 extensions of the identity on K to K.a1 ; a2 /. We now continue in this manner. Theorem 16.4.3. Let LjK be a field extension with M an intermediate field. If a 2 L is separable over K then it is also separable over M .

Section 16.4 Separable Extensions

239

Proof. This follows directly from the fact that the minimal polynomial of a over M divides the minimal polynomial of a over K. Theorem 16.4.4. Let LjK be a field extension. Then the following are equivalent. (1) LjK is finite and separable. (2) There are finitely many separable elements a1 ; : : : ; an over K with K D K.a1 ; : : : ; an /. (3) LjK is finite and if L  L with L algebraically closed then there are exactly ŒL W K monomorphisms ˛ W L ! L with ˛jK D 1K . Proof. That (1) implies (2) follows directly from the definitions. We show then that (2) implies (3). Let L D K.a1 ; : : : ; an / where a1 ; : : : ; an are separable elements over K. The extension LjK is finite (see Theorem 5.3.4). Let pi be the number of pairwise distinct zeros in L of the minimal polynomial mai .x/ D fi .x/ of ai over K.a1 ; : : : ; ai1 /. Then pi  deg.fi / D jK.a1 ; : : : ; ai / W K.a1 ; : : : ; ai 1 /j. Hence pi D deg.fi .x// since ai is separable over K.a1 ; : : : ; ai 1 / from Theorem 16.4.3. Therefore ŒL W K D p1    pn is equal to the number of monomorphisms ˛ W L ! L with ˛jK the identity on K. Finally we show that (3) implies (1). Suppose then the conditions of (3). Since LjK is finite there are finitely many a1 ; : : : ; an 2 L with L D K.a1 ; : : : ; an /. Let pi and fi .x/ be as in the proof above and hence pi  deg.fi .x//. By assumption we have ŒL W K D p1    pn equal to the number of monomorphisms ˛ W L ! L with ˛jK the identity on K. Also ŒL W K D p1    pn  deg.f1 .x//    deg.fn .x// D ŒL W K: Hence pi D deg.fi .x//. Therefore by definition each ai is separable over K. To complete the proof we must show that LjK is separable. Inductively it suffices to prove that K.a1 /jK is separable over K whenever a1 is separable over K and not in K. This is clear if char.K/ D 0 because K is perfect. Suppose then that char.K/ D p p p > 0. First we show that K.a1 / D K.a1 /. Certainly K.a1 /  K.a1 /. Assume that p p p a1 … K.a1 /. Then g.x/ D x  a1 is the minimal polynomial of a1 over K. This p follows from the fact that x p  a1 D .x  a1 /p and hence there can be no irreducible p factor of x p  a1 of the form .x  a1 /m with m < p and mjp. However it follows then in this case that g 0 .x/ D 0 contradicting the separability p of a1 over K. Therefore K.a1 / D K.a1 /. p Let E D K.a1 / then also E D K.E / where E p is the field generated by the p-th powers of E. Now let b 2 E D K.a1 /. We must show that the minimal polynomial of b, say mb .x/, is separable over K.

240

Chapter 16 Separable Field Extensions

Assume that mb .x/ is not separable over K. Then mb .x/ D

k X

bi x pi ;

bi 2 K; bk D 1

i D0

from Theorem 16.2.3. We have b0 C b1 b p C    C bk b pk D 0: Therefore the elements 1; b p ; : : : ; b pk are linearly dependent over K. Since K.a1 / D E D K.E p / we find that 1; b; : : : ; b k are linearly dependent also since if they were independent the p-th powers would also be independent. However this is not possible since k < deg.mb .x//. Therefore mb .x/ is separable over K and hence K.a1 /jK is separable. Altogether LjK is then finite and separable completing the proof. Theorem 16.4.5. Let LjK be a field extension and let M be an intermediate field. Then the following are equivalent. (1) LjK is separable. (2) LjM and M jK are separable. Proof. We first show that (1) ) (2): If LjK is separable then LjM is separable by Theorem 16.4.3 and M jK is separable. Now suppose (2) and let M jK and LjM be separable. Let a 2 L and let ma .x/ D f .x/ D b0 C    C bn1 x n1 C x n be the minimal polynomial of a over M . Then f .x/ is separable. Let M 0 D K.b1 ; : : : ; bn1 /: We have K  M 0  M and hence M 0 jK is separable since M jK is separable. Further a is separable over M 0 since f .x/ is separable and f .x/ 2 M 0 Œx. From Theorem 16.4.1 there are m D deg.f .x// D ŒM 0 .a/ W M 0  extensions of ˛ W M 0 ! M with M the algebraic closure of M 0 . Since M 0 jK is separable and finite there are ŒM 0 W K monomorphisms ˛ W M 0 ! M from Theorem 16.4.4. Altogether there are ŒM 0 .a/ W K monomorphisms ˛ W M 0 ! M with ˛jK the identity on K. Therefore M 0 .a/jK is separable from Theorem 16.4.4. Hence a is separable over K and then LjK is separable. Therefore (2) implies (1). Theorem 16.4.6. Let LjK be a field extension and let S  L such that all elements of S are separable over K. Then K.S /jK is separable and KŒS D K.S /.

Section 16.5 Separability and Galois Extensions

241

Proof. Let W be the set of finite subsets of S. Let T 2 W . From Theorem 16.4.4 we obtain that K.T /jK is separable. Since each element of K.S / is contained in some K.T / we have that K.S /jK is separable. Since all elements of S are algebraic we have that KŒS D K.S /, Theorem 16.4.7. Let LjK be a field extension. Then there exists in L a uniquely determined maximal field M with the property that M jK is separable. If a 2 L is separable over M then a 2 M . M is called the separable hull of K in L. Proof. Let S be the set of all elements in L which are separable over K. Define M D K.S /. Then M jK is separable from Theorem 16.4.6. Now, let a 2 L be separable over M . Then M.a/jM is separable from Theorem 16.4.4. Further M.a/jK is separable from Theorem 16.4.5. It follows that a 2 M .

16.5

Separability and Galois Extensions

We now completely characterize Galois extensions LjK as finite, normal, separable extensions. Theorem 16.5.1. Let LjK be a field extension. Then the following are equivalent. (1) LjK is a Galois extension. (2) L is the splitting field of a separable polynomial in KŒx. (3) LjK is finite, normal and separable. Therefore we may characterize Galois extensions of a field K as finite, normal and separable extensions of K. Proof. Recall from Theorem 8.2.2 that an extension LjK is normal if (1) Ljk is algebraic and (2) each irreducible polynomial f .x/ 2 KŒx that has a zero in L splits into linear factors in LŒx. Now suppose that LjK is a Galois extension. Then LjK is finite from Theorem 15.4.1. Let L D K.b1 ; : : : ; bm / and mbi .x/ D fi .x/ be the minimal polynomial of bi over K. Let ai1 ; : : : ; ain be the pairwise distinct elements from Hi D ¹˛.bi / W ˛ 2 Aut.LjK/º: Define gi .x/ D .x  ai1 /    .x  ain / 2 LŒx: If ˛ 2 Aut.LjK/ then ˛.gi / D gi since ˛ permutes the elements of Hi . This means that the coefficients of gi .x/ are in Fix.L; Aut.LjK// D K. Further gi .x/ 2 KŒx

242

Chapter 16 Separable Field Extensions

because bi is one of the aij and fi .x/jgi .x/. The group Aut.LjK/ acts transitively on ¹ai1 ; : : : ; ain º by the choice of ai1 ; : : : ; ain . Therefore each gi .x/ is irreducible (see Theorem 15.2.4). It follows that fi .x/ D gi .x/. Now fi .x/ has only simple zeros in L, that is no zero has multiplicity  2 and hence fi .x/ splits over L. Therefore L is a splitting field of f .x/ D f1 .x/    fm .x/ and f .x/ is separable by definition. Hence (1) implies (2). Now suppose that L is a splitting field of the separable polynomial f .x/ 2 KŒx and LjK is finite. From Theorem 16.4.4 we get that LjK is separable since L D K.a1 ; : : : ; an / with each ai separable over K. Therefore LjK is normal from Definition 8.2.1. Hence (2) implies (3). Finally suppose that LjK is finite, normal and separable. Since LjK is finite and separable from Theorem 16.4.4 there exist exactly ŒL W K monomorphisms ˛ W L ! L, L the algebraic closure of L, with ˛jK the identity on K. Since LjK is normal these monomorphisms are already automorphisms of L from Theorem 8.2.2. Hence ŒL W K  jAut.LjK/j. Further jL W Kj  jAut.LjK/j from Theorem 15.4.3. Combining these we have ŒL W K D Aut.LjK/ and hence LjK is a Galois extension from Theorem 15.4.9. Therefore (3) implies (1) completing the proof. Recall that any field of characteristic 0 is perfect and therefore any finite extension is separable. Applying this to Q implies that the Galois extensions of the rationals are precisely the splitting fields of polynomials. Corollary 16.5.2. The Galois extensions of the rationals are precisely the splitting fields of polynomials in QŒx. Theorem 16.5.3. Let LjK be a finite, separable field extension. Then there exists an extension field M of L such that M jK is a Galois extension. Proof. Let L D K.a1 ; : : : ; an / with all ai separable over K. Let fi .x/ be the minimal polynomial of ai over K. Then each fi .x/ and hence also f .x/ D f1 .x/    fn .x/ is separable over K. Let M be the splitting field of f .x/ over K. Then M jK is a Galois extension from Theorem 16.5.1. Example 16.5.4. Let K D Q bep the rationals and let f .x/ D x 4  2 2 QŒx. From 4 Chapter 8 we know that L D Q. 2; i/ is a splitting field of f .x/. By the Eisenstein criteria f .x/ is irreducible and ŒL W Q D 8. Moreover p p p p 4 4 4 4 2; i 2;  2; i 2 are the zeros of f .x/. Since the rationals are perfect, f .x/ is separable. LjK is a Galois extension by Theorem 16.5.1. From the calculations in Chapter 15 we have jAut.LjK/j D jAut.L/j D ŒL W K D 8:

243

Section 16.5 Separability and Galois Extensions

Let G D Aut.LjK/ D Aut.LjQ/ D Aut.L/: We want to determine the subgroup lattice of the Galois group G. We show G Š D4 the dihedral group of order 8. Since there are 4 zeros of f .x/ and G permutes these G must be a subgroup of S4 and since the order is 8, G is a 2-Sylow subgroup of S4 . From this we have that G D h.2; 4/; .1; 2; 3; 4/i: If we let  D .2; 4/ and D .1; 2; 3; 4/ we get the isomorphism between G and D4 . From Theorem 14.1.1 we know that D4 D hr; f I r 4 D f 2 D .rf /2 D 1i. This can also be seen in the following manner. Let a1 D

p 4

2;

a2 D i

p 4

2;

p 4 a3 D  2;

a4 D i

p 4

2:

p Let ˛ 2 G. ˛ is determined if we know ˛. 4 2/ and ˛.i /. The possibilities for ˛.i / are i or i that is the zeros of x 2 C 1. p 4 The possibilities for 2 are the 4 zeros of f .x/ D x 4  2. Hence we have 8 possibilities for ˛. These are exactly the elements of the group G. We have ı;  2 G with p p 4 4 ı. 2/ D i 2; ı.i/ D i and

p p 4 4  . 2/ D 2;

 .i/ D i:

It is straightforward to show that ı has order 4,  has order 2 and ı has order 2. These define a group of order 8 isomorphic to D4 and since G has 8 elements this must be all of G. We now look at the subgroup lattice of G and then the corresponding field lattice. Let ı and  be as above. Then G has 5 subgroups of order 2 ¹1; ı 2 º; ¹1;  º; ¹1; ı º; ¹1; ı 2  º; ¹1; ı 3  º: Of these only ¹1; ı 2 º is normal in G. G has 3 subgroups of order 4 ¹1; ı; ı 2 ; ı 3 º; ¹1; ı 2 ; ;  ı 2 º; ¹1; ı 2 ; ı; ı 3  º and all are normal since they all have index 2.

244

Chapter 16 Separable Field Extensions

Hence we have the following subgroup lattice:

From this we construct the lattice of fields and intermediate fields. Since there are 10 proper subgroups of G, from the fundamental theorem of Galois theory there are 10 intermediate fields in LjQ namely the fix fields Fix.L; H / where H is a proper subgroup of G. In the identification the extension field corresponding to the whole group G is the ground field Q (recall that the lattice of fields is the inverted lattice of the subgroups), while the extension field corresponding to the identity is the whole field L. We now consider the other proper subgroups. Let ı;  be as before. p so that (1) Consider M1 D Fix.L; ¹1;  º/. Now ¹1;  º fixes Q. 4 2/ elementwise p p 4 4 Q. 2/  M1 . Further p ŒL W M1  D j¹1;  ºj D 2 and hence ŒL W Q. 2/ D 2. Therefore M1 D Q. 4 2/. (2) Consider M2 D Fix.L; ¹1;  ıº/. We have p p p 4 4 4  ı. 2/ D  .i 2/ D i 2 p p p 4 4 4  ı.i 2/ D  . 2/ D  2 p p p 4 4 4  ı. 2/ D  .i 2/ D i 2 p p p 4 4 4  ı.i 2/ D  . 2/ D 2: p p It follows that  ı fixes .1  i / 4 2 and hence M2 D Q..1  i / 4 2/. (3) Consider M3 D Fix.L; ¹1;  ı2 º. p The map  ı 2 interchanges a1 and a3 and fixes a2 and a4 . Therefore M3 D Q.i 4 2/.

245

Section 16.6 The Primitive Element Theorem

In an analogous manner we can then consider the other p 5 proper subgroups p and corresponding intermediate fields. If we let b1 D .1  i / 4 2 and b2 D .1 C i / 4 2 we get the following lattice of fields and subfields.

16.6

The Primitive Element Theorem

In this section we describe finite separable field extensions as simple extensions. If follows that a Galois extension is always a simple extension. Theorem 16.6.1 (primitive element theorem). Let L D K. 1 ; : : : ; n / and suppose that each i is separable over K. Then there exists a 0 2 L such that L D K. 0 /. The element 0 is called a primitive element. Proof. Suppose first that K is a finite field. Then L is also a finite field and therefore L? D h 0 i is cyclic. Therefore L D K. 0 / and the theorem is proved if K is a finite field. Now suppose that K is infinite. Inductively it suffices to prove the theorem for n D 2. Hence let ˛; ˇ 2 L be separable over K. We must show that there exists a

2 L with K.˛; ˇ/ D K. /. Let L be the splitting field of the polynomial m˛ .x/mˇ .x/ over L where m˛ .x/; mˇ .x/ are respectively the minimal polynomials of ˛; ˇ over K. In LŒx we have m˛ .x/ D .x  ˛1 /.x  ˛2 /    .x  ˛s / with ˛ D ˛1 mˇ .x/ D .x  ˇ1 /.x  ˇ2 /    .x  ˇ t /

with ˇ D ˇ1 :

By assumption the ˛i and the ˇj are respectively pairwise distinct. For each pair .i; j / with 1  i  s, 2  j  t the equation ˛1 C zˇ1 D ˛i C zˇj

246

Chapter 16 Separable Field Extensions

has exactly one solution z 2 L since ˇj  ˇ1 ¤ 0 if j  2. Since K is infinite there exists a c 2 K with ˛1 C cˇ1 ¤ ˛i C cˇj for all i; j with 1  i  s, 2  j  t . With such a value c 2 K we define

D ˛ C cˇ D ˛1 C cˇ1 : We claim that K.˛; ˇ/ D K. /. It suffices to show that ˇ 2 K. / for then ˛ D

 cˇ 2 K. /. This implies that K.˛; ˇ/  K. / and since 2 K.˛; ˇ/ it follows that K.˛; ˇ/ D K. /. To show that ˇ 2 K. / we first define f .x/ D m˛ .  cx/ and let d.x/ D gcd.f .x/; mˇ .x//. We may assume that d.x/ is monic. We show that d.x/ D x  ˇ. Then ˇ 2 K. / since d.x/ 2 K. /Œx. Assume first that d.x/ D 1. Then gcd.f .x/; mˇ .x// D 1 and f .x/ and mˇ .x/ are also relatively prime in LŒx. This is a contradiction since f .x/ and mˇ .x/ have the common zero ˇ 2 L and hence the common divisor x  ˇ. Therefore d.x/ ¤ 1 so deg.d.x//  1. The polynomial d.x/ is a divisor of mˇ .x/ and hence d.x/ splits into linear factors of the form x  ˇj , 1  j  t in LŒx. The proof is completed if we can show that no linear factor of the form x  ˇj with 2  j  t is a divisor of f .x/. That is, we must show that f .ˇj / ¤ 0 in L if j  2. Now f .ˇj / D m˛ .  cˇj / D m˛ .˛1 C cˇ1  cˇj /. Suppose that f .ˇj / D 0 for some j  2. This would imply that ˛i D ˛1 Ccˇ1 cˇj , that is, ˛1 Ccˇ1 D ˛j Ccˇj for j  2. This contradicts the choice of the value c. Therefore f .ˇj / ¤ 0 if j  2 completing the proof. In the above theorem it is sufficient to assume that n  1 of 1 ; : : : ; n are separable over K. The proof is similar. We only need that the ˇ1 ; : : : ; ˇ t are pairwise distinct if ˇ is separable over K to show that K.˛; ˇ/ D K. / for some 2 L. If K is a perfect field then every finite extension is separable. Therefore we get the following corollary. Corollary 16.6.2. Let LjK be a finite extension with K a perfect field. Then L D K. / for some 2 L. Corollary 16.6.3. Let LjK be a finite extension with K a perfect field. Then there exist only finitely many intermediate fields E with K  E  L. Proof. Since K is a perfect field we have L D K. / for some 2 L. Let m .x/ 2 KŒx be the minimal polynomial of over K and let L be the splitting field of m .x/ over K. Then LjK is a Galois extension and hence there are only finitely many intermediate fields between K and L. Therefore also only finitely many between K and L.

Section 16.7 Exercises

247

Suppose that LjK is algebraic. Then in general L D K. / for some 2 L if and only if there exist only finitely many intermediate fields E with K  E  L. This condition on intermediate fields implies that LjK is finite if LjK is algebraic. Hence we have proved this result in the case that K is perfect. The general case is discussed in the book of S. Lang [8].

16.7

Exercises

1. Let f .x/ D x 4  8x 3 C 24x 2  32x C 14 2 QŒx and let v 2 C be a zero of f . Let ˛ WD v.4  v/ and K a splitting field of f over Q. Show: (i) f is irreducible over Q and f .x/ D f .4  x/. (ii) There is exactly one automorphism of Q.v/ with .v/ D 4  v. (iii) L WD Q.˛/ is the Fix field of and jL W Qj D 2. (iv) Determine the minimal polynomial of ˛ over Q and determine ˛. (v) jQ.v/ W Lj D 2 and determine the minimal polynomial of v over L and determine v and all other zeros of f .x/. (vi) Determine the degree of jK W Qj. (vii) Determine the structure of Aut.KjQ/. 2. Let LjK be a field extension and f 2 KŒx a separable polynomial. Let Z be a splitting field of f over L and Z0 a splitting field of f over K. Show that Aut.ZjL/ is isomorphic to a subgroup of Aut.Z0 jK/. 3. Let LjK be a field extension and v 2 L. For each element c 2 K it is K.v C c/ D K.v/. For c ¤ 0 it is K.cv/ D K.v/. p p p p presentable as a 4. Let v D 2 C 3 and let K D Q.v/. Show that 2 and 3parep Q-linear combination of 1; v; v 2 ; v 3 . Conclude that K D Q. 2; 3/. 5. Let L be the splitting field of x 3  5 over Q in C. Determine a primitive element t of L over Q.

Chapter 17

Applications of Galois Theory

17.1

Applications of Galois Theory

As we mentioned in Chapter 1 Galois theory was originally developed as part of the proof that polynomial equations of degree 5 or higher over the rationals cannot be solved by formulas in terms of radicals. In this chapter we do this first and prove the insolvability of the quintic by radicals. To do this we must examine in detail what we call radical extensions. We then return to some geometric material we started in Chapter 6. There using general field extensions we proved the impossibility of certain geometric compass and straightedge constructions. Here we use Galois theory to consider constructible n-gons. Finally we will use Galois theory to present a proof of the fundamental theorem of algebra which says essentially that the complex number field C is algebraically closed.

17.2

Field Extensions by Radicals

We would like to use Galois theory to prove the insolvability by radicals of polynomial equations of degree 5 or higher. To do this we must introduce extensions by radicals and solvability by radicals. Definition 17.2.1. Let LjK be a field extension. (1) Each zero of a polynomial x n  a 2 KŒx in L is called a radical (over K). We p denote it by n a (if a more detailed identification is not necessary). p (2) L is called a simple extension of K by a radical if L D K. n a/ for some a 2 K. (3) L is called an extension of K by radicals if there is a chain of fields K D L 0  L 1      Lm D L such that each Li is a simple extension of Li 1 by a radical for each i D 1; : : : ; m. (4) Let f .x/ 2 KŒx. Then the equation f .x/ D 0 is solvable by radicals or just solvable if the splitting field of f .x/ over K is contained in an extension of K by radicals.

249

Section 17.2 Field Extensions by Radicals

In proving the insolvability of the quintic we will look for necessary and sufficient conditions for the solvability of polynomial equations. Our main result will be that if f .x/ 2 KŒx then f .x/ D 0 is solvable over K if the Galois group of the splitting field of f .x/ over K is a solvable group (see Chapter 11). In the remainder of this section we assume that all fields have characteristic zero. The next theorem gives a characterization of simple extensions by radicals. Theorem 17.2.2. Let LjK be a field extension and n 2 N. Assume that the polynomial x n  1 splits into linear factors in KŒx so that K contains all the n-th roots of p unity. Then L D K. n a/ for some a 2 K if and only if L is a Galois extension over K and Aut.LjK/ D Z=mZ for some m 2 N with mjn. Proof. The n-th roots of unity, that is the zeros of the polynomial x n  1 2 KŒx, form a cyclic multiplicative group F  K ? of order n since each finite subgroup of the multiplicative group K ? of K is cyclic and jF j D n. We call an n-th root of unity ! primitive if F D h!i. p Now let L D K. n a/ with a 2 K, that is, L D K.ˇ/ with ˇ n D a 2 K. Let ! be a primitive n-th root of unity. With this ˇ the elements !ˇ; ! 2 ˇ; : : : ; ! n ˇ D ˇ are zeros of x n  a. Hence the polynomial x n  a splits into linear factors over L and hence L D K.ˇ/ is a splitting field of x n  a over K. It follows that LjK is a Galois extension. Let 2 Aut.LjK/. Then .ˇ/ D !  ˇ for some 0 <   n. The element !  is uniquely determined by and we may write !  D ! . Consider the map  W Aut.LjK/ ! F given by ! ! where ! is defined as above by .ˇ/ D ! ˇ. If ; 2 Aut.LjK/ then

 .ˇ/ D .! / .ˇ/ D ! ! ˇ because ! 2 K. Therefore .  / D . /. / and hence  is a homomorphism. The kernel ker./ contains all the K-automorphisms of L for which .ˇ/ D ˇ. However since K D K.ˇ/ it follows that ker./ contains only the identity. The Galois group Aut.LjK/ is therefore isomorphic to a subgroup of F . Since F is cyclic of order n we have that Aut.LjK/ is cyclic of order m for some mjn completing one way in the theorem. Conversely first suppose that LjK is a Galois extension with Aut.LjK/ D Zn a cyclic group of order n. Let be a generator of Aut.LjK/. This is equivalent to Aut.LjK/ D ¹ ; 2 ; : : : ; n D 1º: Let ! be a primitive n-th root of unity. Then by assumption ! 2 K, .!/ D ! and F D ¹!; ! 2 ; : : : ; ! n D 1º. Further the pairwise distinct automorphism  ,  D 1; 2; : : : ; n, of L are linearly independent, that is there exists an  2 L such that !?D

n X D1

!   ./ ¤ 0:

250

Chapter 17 Applications of Galois Theory

The element ! ?  is called the Lagrange resolvent of ! by . We fix such an element  2 L. Then we get, since .!/ D !,

.! ? / D

n X

 C1

!

D1

D ! 1

./ D !

1

n X

!

C1 C1

./ D !

D1 n X

1

nC1 X

!   ./

D2

!   ./ D ! 1 .! ? /:

D1

Further  .! ? / D !  .! ? /,  D 1; 2; : : : ; n. Hence the only K-automorphism of L which fixes ! ?  is the identity. Therefore Aut.LjK.! ? // D ¹1º and hence L D K.! ? / by the fundamental theorem of Galois theory. Further

..! ? /n / D . .! ? //n D .! 1 .! ? //n D ! n .! ? /n D .! ? /n : Therefore .! ? /n 2 Fix.L; Aut.LjK// D K again from the fundamental theorem p of Galois theory. If a D .! ? /n 2 K then first a 2 K and second L D K. n a/ D K.! ? /. This proves the result in the case where m D n. We now use this to prove it in general. Suppose that LjK is a Galois extension with Aut.LjK/ D Zm a cyclic p group of m order m where n D q m for some q  1. If n D q m then L D K. b/ for some b 2 K by the above argument. Hence L D K.ˇ/ with ˇ m 2 K. Then certainly p a D ˇ n D .ˇ m /q 2 K and therefore L D K.ˇ/ D K. n a/ for some a 2 K completing the general case. We next show that every extension by radicals is contained in a Galois extension by radicals. Theorem 17.2.3. Each extension L of K by radicals is contained in a Galois extension LQ of K by radicals. This means that there is an extension LQ of K by radicals Q with L  LQ and LjK is a Galois extension. Proof. We use induction on the degree m D ŒL W K. Suppose that m D 1. If p L D K. n a/ then if ! is a primitive n-th root of unity define KQ D K.!/ and LQ D p Q Q n a/. We then get the chain K  KQ  LQ with L  LQ and LjK is a Galois K. extension. This last statement is due to the fact that LQ is the splitting field of the polynomial x n  a 2 KŒx over K. Hence the theorem is true if m D 1. Now suppose that m  2 and suppose that the theorem is true for all extensions F of K by radicals with ŒF W K < m. Since m  2 by the definition of extension by radicals there exists a simple extension LjE by a radical. That is there exists a field E with K  E  L;

ŒL W E  2

Section 17.2 Field Extensions by Radicals

251

p and L D E. n a/ for some a 2 E; n 2 N. Now ŒE W K < m so be the inductive Q Let G D hypothesis there exists a Galois extension by radicals EQ of K with E  E. Q Aut.EjK/ and let LQ be the splitting field of the polynomial f .x/ D ma .x n / 2 KŒx over EQ where ma .x/ is the minimal polynomial of a over K. We show that LQ has the desired properties. p Q Therefore LQ Now n a 2 L is a zero of the polynomial f .x/ and E  EQ  L. p n contains an E-isomorphic image of L D K. a/ and hence we may consider LQ as an extension of L. Since EQ is a Galois extension of K the polynomial f .x/ may be factored as f .x/ D .x n  ˛1 /    .x n  ˛s / Q Therefore with ˛i 2 EQ for i D 1; : : : ; s. All zeros of f .x/ in LQ are radicals over E. Q Q Q L is an extension by radicals of E. Since E is also an extension by radicals of K we obtain that LQ is an extension by radicals of K. Since EQ is a Galois extension of K we have that EQ is a splitting field of a polynomial Q is a splitting field of f .x/ 2 KŒx over E. Q Altogether then g.x/ 2 KŒx. Further L Q we have that L is a splitting field of f .x/g.x/ 2 KŒx over K. Therefore LQ is a Galois extension of K completing the proof. We will eventually show that a polynomial equation is solvable by radicals if and only if the corresponding Galois group is a solvable group. We now begin to find conditions where the Galois group is solvable. Lemma 17.2.4. Let K D L0  L1      Lr D L be a chain of fields such that the following hold: (i) L is a Galois extension of K. (ii) Lj is a Galois extension of Lj 1 for j D 1; : : : ; r. (iii) Gj D Aut.Lj jLj 1 / is abelian for j D 1; : : : ; r. Then G D Aut.LjK/ is solvable. Proof. We prove the lemma by induction on r. If r D 0 then G D ¹1º and there is nothing to prove. Suppose then that r  1 and assume that the lemma holds for all such chains of fields with a length r 0 < r. Since L1 jK is a Galois extension then Aut.L1 jK/ is a normal subgroup of G by the fundamental theorem of Galois theory and further G1 D Aut.L1 jK/ D G= Aut.LjL1 /: Since G1 is an abelian group it is solvable and by assumption Aut.LjL1 / is solvable. Therefore G is solvable (see Theorem 12.2.4). Lemma 17.2.5. Let LjK be a field extension. Let KQ and LQ be the splitting fields of the polynomial x n  1 2 KŒx over K and L respectively. Since K  L we have Q Then the following hold: KQ  L.

252

Chapter 17 Applications of Galois Theory

Q Q (1) If 2 Aut.LjL/ then jKQ 2 Aut.KjK/ and the map Q Q Aut.LjL/ ! Aut.KjK/

given by 7! jKQ

is an injective homomorphism. Q (2) Suppose that in addition LjK is a Galois extension. Then LjK is also a Galois Q Q extension. If further 2 Aut.LjK/ then jL 2 Aut.LjK/ and Q K/ Q ! Aut.LjK/ given by 7! jL Aut.Lj is an injective homomorphism. Proof. (1) Let ! be a primitive n-th root of unity. Then KQ D K.!/ and LQ D L.!/. Q Each 2 Aut.LjL/ maps ! onto a primitive n-th root of unity and fixes K  L Q Q Certainly elementwise. Hence from 2 Aut.LjL/ we get that jKQ 2 Aut.KjK/. Q Q the map 7! jKQ defines a homomorphism Aut.LjL/ ! Aut.KjK/. Let jKQ D 1 Q with 2 Aut.LjL/. Then .!/ D ! and therefore we have already that D 1 since Q L D L.!/. (2) If L is the splitting field of a polynomial g.x/ over K then LQ is the splitting field Q of g.x/.x n  1/ over K. Hence LjK is a Galois extension. Therefore K  L  LQ Q Q and LjK; LjL and LjK are all Galois extensions. Therefore from the fundamental theorem of Galois theory Q Aut.LjK/ D ¹ jL I 2 Aut.LjK/º: Q K/. Q Certainly the map Aut.Lj Q K/ Q ! In particular jL 2 Aut.LjK/ if 2 Aut.Lj Q K/ Q we get that Aut.LjK/ given by 7! jL is a homomorphism. From 2 Aut.Lj

.!/ D ! where as above ! is a primitive n-th root of unity. Therefore if jL D 1 Q D L.!/. Hence the map is injective. then already D 1 since L

17.3

Cyclotomic Extensions

Very important in the solvability by radicals problem are the splitting fields of the polynomials x n  1 over Q. These are called cyclotomic fields. Definition 17.3.1. The splitting field of the polynomial x n  1 2 QŒx with n  2 is called the n-th cyclotomic field denoted kn . We have kn D Q.!/ where ! is a primitive n-th root of unity, for example ! D over Q. kn jQ is a Galois extension and the Galois group Aut.kn jQ/ is the set e of automorphisms m W ! ! ! m with 1  m  n and gcd.m; n/ D 1. To understand this group G we need the following concept. A prime residue class mod n is a residue class a C nZ with gcd.a; n/ D 1. The set of the prime residue 2 i n

Section 17.4 Solvability and Galois Extensions

253

classes mod n is just the set of invertible elements with respect to multiplication of the Z=nZ. This forms a multiplicative group that we denote by .Z=nZ/? D Pn . We have jPn j D .n/ where .n/ is the Euler phi-function. If G D Aut.kn jQ/ then clearly G Š Pn under the map m 7! m C nZ. If n D p is a prime number then G D Aut.kn jQ/ is cyclic with jGj D p  1. If n D p 2 then jGj D jAut.kp 2 jQ/j D p.p  1/ since x p 1 x  1 D x p.p1/ C x p.p1/1 C    C 1 x  1 xp  1 2

and each primitive p-th root of unity is a zero of this polynomial. Lemma 17.3.2. Let K be a field and KQ be the splitting field of x n  1 over K. Then Q Aut.KjK/ is abelian. Proof. We apply Lemma 17.2.5 for the field extension KjQ. This can be done since Q the characteristic of K is zero and Q is the prime field of K. It follows that Aut.KjK/ Q Q D kn is isomorphic to a subgroup of Aut.QjQ/ from part (1) of Lemma 17.2.5. But Q Q Q and hence Aut.QjQ/ is abelian. Therefore Aut.KjK/ is abelian.

17.4

Solvability and Galois Extensions

In this section we prove that solvability by radicals is equivalent to the solvability of the Galois group. Theorem 17.4.1. Let LjK be a Galois extension of K by radicals. Then G D Aut.LjK/ is a solvable group. Proof. Suppose that LjK is a Galois extension. Then we have a chain of fields K D L0  L1      Lr D L p such that Lj D Lj 1 . nj aj / for some aj 2 Lj . Let n D n1    nr and let LQ j be the splitting field of the polynomial x n  1 2 KŒx over Lj for each j D 0; 1; : : : ; r. p Q j 1 . nj aj / and we get the chain Qj D L Then L Q0  L Q 1      LQ r D L: Q K  KQ D L Q From part (2) of Lemma 17.2.5 we get that LjK is a Galois extension. Further Q Q Q Q Lj jLj 1 is a Galois extension with Aut.Lj jLj 1 / cyclic from Theorem 17.2.2. EspeQ j 1 / is abelian. The group Aut.KjK/ Q Q j jL is abelian from Lemma 17.3.2. cially Aut.L Therefore we may apply Lemma 17.2.4 to the chain Q 0      LQ r D L: Q K  KQ D L

254

Chapter 17 Applications of Galois Theory

Q Therefore GQ D Aut.LjK/ is solvable. The group G D Aut.LjK/ is a homomorphic Q image of G from the fundamental theorem of Galois theory. Since homomorphic images of solvable groups are still solvable (see Theorem 12.2.3) it follows that G is solvable. Lemma 17.4.2. Let LjK be a Galois extension and suppose that G D Aut.LjK/ is solvable. Assume further that K contains all q-th roots of unity for each prime divisor q of m D ŒL W K. Then L is an extension of K by radicals. Proof. Let LjK be a Galois extension and suppose that G D Aut.LjK/ is solvable and assume that K contains all the q-th roots of unity for each prime divisor q of m D ŒL W K. We prove the result by induction on m. If m D 1 then L D K and the result is clear. Now suppose that m  2 and assume that the result holds for all Galois extensions L0 jK 0 with ŒL0 W K 0  < m. Now G D Aut.LjK/ is solvable and G is nontrivial since m  2. Let q be a prime divisor of m. From Lemma 12.2.2 and Theorem 13.3.5 it follows that there is a normal subgroup H of G with G=H cyclic of order q. Let E D Fix.L; H /. From the fundamental theorem of Galois theory EjK is a Galois extension with Aut.EjK/ Š G=H and hence Aut.EjK/ is cyclic of order q. From Theorem 17.2.2 EjK is a simple extension of K by a radical. The proof is completed if we can show that L is an extension of E by radicals. The extension LjE is a Galois extension and the group Aut.LjE/ is solvable since it is a subgroup of G D Aut.LjK/. Each prime divisor p of ŒL W E is also a prime divisor of m D ŒL W K by the degree formula. Hence as an extension of K the field E contains all the p-th roots of unity. Finally ŒL W E D

m ŒL W K D < m: ŒE W K q

Therefore LjE is an extension of E by radicals from the inductive assumption completing the proof.

17.5

The Insolvability of the Quintic

We are now able to prove the insolvability of the quintic. This is one of the most important applications of Galois theory. As we mentioned we do this by equating the solvability of a polynomial equation by radicals to the solvability of the Galois group of the splitting field of this polynomial. Theorem 17.5.1. Let K be a field of characteristic 0 and let f .x/ 2 KŒx. Suppose that L is the splitting field of f .x/ over K. Then the polynomial equation f .x/ D 0 is solvable by radicals if and only if Aut.LjK/ is solvable.

Section 17.5 The Insolvability of the Quintic

255

Proof. Suppose first that f .x/ D 0 is solvable by radicals. Then L is contained in an extension L0 of K by radicals. Hence L is contained in a Galois extension LQ of Q K by radicals from Theorem 17.2.3. The group GQ D Aut.LjK/ is solvable from Theorem 17.4.1. Further LjK is a Galois extension. Therefore the Galois group Q Aut.LjK/ is solvable as a subgroup of G. Conversely suppose that the group Aut.LjK/ is solvable. Let q1 ; : : : ; qr be the prime divisors of m D ŒK W K and let n D q1    qr . Let KQ and LQ be the splitting Q fields of the polynomial x n  1 2 KŒx over K and L respectively. We have KQ  L. Q Q Q From part (2) of Lemma 17.2.5 we have that LjK is a Galois extension and Aut.LjK/ Q D is isomorphic to a subgroup of Aut.LjK/. From this we first obtain that ŒLQ W K Q Q jAut.LjK/j is a divisor of ŒL W K D jAut.LjK/j. Hence each prime divisor q of Q is also a prime divisor of ŒL W K. Therefore LQ is an extension by radicals ŒLQ W K of KQ by Lemma 17.4.2. Since KQ D K.!/ where ! is a primitive n-th root of unity we obtain that LQ is also an extension of K by radicals. Therefore L is contained in an extension LQ of K by radicals and therefore f .x/ D 0 is solvable by radicals. Corollary 17.5.2. Let K be a field of characteristic 0 and let f .x/ 2 KŒx be a polynomial of degree m with 1  m  4. Then the equation f .x/ D 0 is solvable by radicals. Proof. Let L be the splitting field of f .x/ over K. The Galois group Aut.LjK/ is isomorphic to the subgroup of the symmetric group Sm . Now the group S4 is solvable via the chain ¹1º  Z2  D2  A4  S4 where Z2 is the cyclic group of order 2 and D2 is the Klein 4-group which is isomorphic to Z2  Z2 . Because Sm  S4 for 1  m  4 it follows that Aut.LjK/ is solvable. From Theorem 17.5.1 the equation f .x/ D 0 is solvable by radicals. Corollary 17.5.2 uses the general theory to show that any polynomial equation of degree less than or equal to 4 is solvable by radicals. This however does not provide explicit formulas for the solutions. We present these below. Let K be a field of characteristic 0 and let f .x/ 2 KŒx be a polynomial of degree m with 1  m  4. Case (1): If deg.f .x// D 1 then f .x/ D ax C b with a; b 2 K and a ¤ 0. A zero is then given by k D  ab . Case (2): If deg.f .x// D 2 then f .x/ D ax 2 C bx C c with a; b; c 2 K and a ¤ 0. The zeros are then given by the quadratic formula p b ˙ b 2  4ac : kD 2a We note that the quadratic formula holds over any field of characteristic not equal to 2. Whether there is a solution within the field K then depends on whether b 2  4ac has a square root within K.

256

Chapter 17 Applications of Galois Theory

For the cases of degrees 3 and 4 we have the general forms of what are known as Cardano’s formulas. Case (3): If deg.f .x// D 3 then f .x/ D ax 3 C bx 2 C cx C d with a; b; c; d 2 K and a ¤ 0. Dividing through by a we may assume without loss of generality that a D 1. By a substitution x D y  b3 the polynomial is transformed into g.y/ D y 3 C py C q 2 KŒy: Let L be the splitting field of g.y/ over K and let ˛ 2 L be a zero of g.y/ so that ˛ 3 C p˛ C q D 0: If p D 0 then ˛ D

p 3

q so that g.y/ has the three zeros p 3

q;

!

p 3

q;

!2

p 3

q

where ! is a primitive third root of unity, ! 3 D 1 with ! ¤ ! 2 . Now let p ¤ 0 and let ˇ be a zero of x 2  ˛x  p3 in a suitable extension L0 of L. p . Putting this into the transformed We have ˇ ¤ 0 since p ¤ 0. Hence ˛ D ˇ  3ˇ cubic equation ˛ 3 C p˛ C q D 0 we get ˇ3 

p3 C q D 0: 27ˇ 3

Define D ˇ 3 and ı D . p /3 so that 3ˇ

C ı C q D 0: Then  3 p D 0 and

C q  3 2

p3  C ı C q D 0 and 27ı

Hence the zeros of the polynomial  3 p x C qx  3 2

are

s  2  3 p q q C :

; ı D  ˙ 2 2 3

 3 p D 0: ı C qı  3 2

257

Section 17.5 The Insolvability of the Quintic

If we have D ı then both are equal to  q2 and s  2  3 p q C D 0: 2 3 Then from the definitions of ; ı we have D ˇ 3 and ı D . p /3 . From above 3ˇ p ˛ D ˇ  3ˇ . Therefore we get ˛ by finding the cube roots of and ı. There are certain possibilities and combinations with these cube roots but because of the conditions the cube roots of and ı are not independent. We must satisfy the condition p p p p 3

3 ıDˇ D : 3ˇ 3 Therefore we get the final result: The zeros of g.y/ D y 3 C py C q with p ¤ 0 are u C v;

!u C ! 2 v;

! 2 u C !v

where ! is a primitive third root of unity and v s u  2  3 u q p q t C and v D uD 3  C 2 2 3

v s u  2  3 u q p q t C :  3  2 2 3

The above is known as the cubic formula or Cardano’s formula. Case (4): If deg.f .x// D 4 then f .x/ D ax 4 C bx 3 C cx 2 C dx C e with a; b; c; d; e 2 K and a ¤ 0. Dividing through by a we may assume without loss of generality that a D 1. b By a substitution x D y  4a the polynomial f .x/ is transformed into g.y/ D y 4 C py 2 C qy C r: The zeros of g.y/ are p p 1 p y1 D . ˛1 C ˛2 C ˛3 / 2 p p 1 p y2 D . ˛1  ˛2  ˛3 / 2 p p 1 p y3 D . ˛1 C ˛2  ˛3 / 2 p p 1 p y4 D . ˛1  ˛2 C ˛3 / 2 where ˛1 ; ˛2 ; ˛3 are the zeros of the cubic polynomial h.z/ D z 3 C 2pz C .p 2  4r/z  q 2 2 QŒz:

258

Chapter 17 Applications of Galois Theory

The polynomial h.z/ is called the cubic resolvent of g.y/. For a detailed proof of the case where m D 4 see [8]. The following theorem is due to Abel and shows the insolvability of the general degree 5 polynomial over the rationals Q. Theorem 17.5.3. Let L be the splitting field of the polynomial f .x/ D x 5 2x 4 C2 2 QŒx over Q. Then Aut.LjK/ D S5 the symmetric group on 5 letters. Since S5 is not solvable the equation f .x/ D 0 is not solvable by radicals. Proof. The polynomial f .x/ is irreducible over Q by the Eisenstein criterion. Further f .x/ has five zeros in the complex numbers C by the fundamental theorem of algebra (see Section 17.7). We claim that f .x/ has exactly 3 real zeros and 2 nonreal zeros which then necessarily are complex conjugates. In particular the 5 zeros are pairwise distinct. To see the claim notice first that f .x/ has at least 3 real zeros from the intermediate value theorem. As a real function f .x/ is continuous and f .1/ D 1 < 0 and f .0/ D 2 > 0 so it must have a real zero between 1 and 0. Further f . 32 / D  81 3 < 0 and f .2/ D 2 > 0. Hence there must be distinct real zeros between 0 and 32 and between 32 and 2. Suppose that f .x/ has more than 3 real zeros. Then f 0 .x/ D x 3 .5x  8/ has at least 3 pairwise distinct real zeros from Rolle’s theorem. But f 0 .x/ clearly has only 2 real zeros so this is not the case. Therefore f .x/ has exactly 3 real zeros and hence 2 nonreal zeros that are complex conjugates. Let L be the splitting field of f .x/. The field L lies in C and the restriction of the map ı W z 7! z of C to L maps the set of zeros of f .x/ onto themselves. Therefore ı is an automorphism of L. The map ı fixes the 3 real zeros and transposes the 2 nonreal zeros. From this we now show that Aut.LjQ/ D Aut L D G D S5 the full symmetric group on 5 symbols. Clearly G  S5 since G acts as a permutation group on the 5 zeros of f .x/. Since ı transposes the 2 nonreal roots, G (as a permutation group) contains at least one transposition. Since f .x/ is irreducible G acts transitively on the zeros of f .x/. Let x0 be one of the zeros of f .x/ and let Gx0 be the stabilizer of x0 . Since G acts transitively x0 has five images under G and therefore the index of the stabilizer must be 5 (see Chapter 10). 5 D ŒG W Gx0  which by Lagrange’s theorem must divide the order of G. Therefore from the Sylow theorems G contains an element of order 5. Hence G contains a 5-cycle and a transposition and therefore by Theorem 11.4.3 it follows that G D S5 . Since S5 is not solvable it follows that f .x/ cannot be solved by radicals. Since Abel’s theorem shows that there exists a degree 5 polynomial that cannot be solved by radicals it follows that there can be no formula like Cardano’s formula in terms of radicals for degree 5.

Section 17.6 Constructibility of Regular n-Gons

259

Corollary 17.5.4. There is no general formula for solving by radicals a fifth degree polynomial over the rationals. We now show that this result can be further extended to any degree greater than 5. Theorem 17.5.5. For each n  5 there exist polynomials f .x/ 2 QŒx of degree n for which the equation f .x/ D 0 is not solvable by radicals. Proof. Let f .x/ D x n5 .x 5 2x 4 C2/ and let L be the splitting field of f .x/ over Q. Then Aut.LjQ/ D Aut.L/ contains a subgroup that is isomorphic to S5 . It follows that Aut.L/ is not solvable and therefore the equation f .x/ D 0 is not solvable by radicals. This immediately implies the following. Corollary 17.5.6. There is no general formula for solving by radicals polynomial equations over the rationals of degree 5 or greater.

17.6

Constructibility of Regular n-Gons

In Chapter 6 we considered certain geometric material related to field extensions. There, using general field extensions, we proved the impossibility of certain geometric compass and straightedge constructions. In particular there were four famous insolvable (to the Greeks) construction problems. The first is the squaring of the circle. This problem is, given a circle, to construct using straightedge and compass a square having area equal to that of the given circle. The second is the doubling of the cube. This problem is given a cube of given side length, to construct, using a straightedge and compass, a side of a cube having double the volume of the original cube. The third problem is the trisection of an angle. This problem is to trisect a given angle using only a straightedge and compass. The final problem is the construction of a regular n-gon. This problems asks which regular n-gons could be constructed using only straightedge and compass. In Chapter 6 we proved the impossibility of the first 3 problems. Here we use Galois theory to consider constructible n-gons. Recall that a Fermat number is a positive integer of the form n

Fn D 22 C 1;

n D 0; 1; 2; 3; : : : :

If a particular Fm is prime it is called a Fermat prime. Fermat believed that all the numbers in this sequence were primes. In fact F0 ; F1 ; F2 ; F3 ; F4 are all prime but F5 is composite and divisible by 641 (see exercises). It is still an open question whether or not there are infinitely many Fermat primes. It has been conjectured that there are only finitely many. On the other hand if a number of

260

Chapter 17 Applications of Galois Theory

the form 2n C 1 is a prime for some integer n then it must be a Fermat prime that is n must be a power of 2. We first need the following. Theorem 17.6.1. Let p D 2n C 1, n D 2s with s  0 be a Fermat prime. Then there exists a chain of fields Q D L0  L1      Ln D kp where kp is the p-th cyclotomic field such that ŒLj W Lj 1  D 2 for j D 1; : : : ; n. Proof. The extension kp jQ is a Galois extension and Œkp W Q D p  1. Further Aut.kp / is cyclic of order p  1 D 2n . Hence there is a chain of subgroups ¹1º D Un  Un1      U0 D Aut.kp / with ŒUj 1 W Uj  D 2 for j D 1; : : : ; n. From the fundamental theorem of Galois theory the fields Lj D Fix.kp ; Uj / with j D 0; : : : ; n have the desired properties. The following corollaries describe completely the constructible n-gons tying them to Fermat primes. Corollary 17.6.2. Consider the numbers 0; 1, that is a unit line segment or a unit circle. A regular p-gon with p  3 prime is constructible from ¹0; 1º using a straights edge and compass if and only if p D 22 C 1; s  0 is a Fermat prime. Proof. From Theorem 6.3.13 we have that if a regular p-gon is constructible with a straightedge and compass then p must be a Fermat prime. The sufficiency follows from Theorem 17.6.1. We now extend this to general n-gons. Let m; n 2 N. Assume that we may construct from ¹0; 1º a regular n-gon and a regular m-gon. In particular this means that 2 2 2 we may construct the real numbers cos. 2 n /; sin. n /; cos. m / and sin. m /. If the gcd.m; n/ D 1 then we may construct from ¹0; 1º a regular mn-gon. To see this notice that             2 2.n C m/ 2 2 2 2 2 C D cos D cos cos  sin sin cos n m nm n m n m and             2 2.n C m/ 2 2 2 2 2 sin D sin D sin cos C cos sin : C n m nm n m n m

Section 17.7 The Fundamental Theorem of Algebra

261

2 2 Therefore we may construct from ¹0; 1º the numbers cos. mn / and sin. mn / because gcd.n C m; mn/ D 1. Therefore we may construct from ¹0; 1º a regular mn-gon. Now let p  3 be a prime. Then Œkp2 W Q D p.p  1/ which is not a power of 2. Therefore from ¹0; 1º it is not possible to construct a regular p 2 -gon. Hence altogether we have the following.

Corollary 17.6.3. Consider the numbers 0; 1, that is a unit line segment or a unit circle. A regular n-gon with n 2 N is constructible from ¹0; 1º using a straightedge and compass if and only if (i) n D 2m , m  0 or (ii) p D 2m p1 p2    pr , m  0 and the pi are pairwise distinct Fermat primes. Proof. Certainly we may construct a 2m -gon. Further if r; s 2 N with gcd.r; s/ D 1 and if we can construct a regular rs-gon then clearly we may construct a regular r-gon and a regular s-gon.

17.7

The Fundamental Theorem of Algebra

The fundamental theorem of algebra is one of the most important algebraic results. This says that any nonconstant complex polynomial must have a complex zero. In the language of field extensions this says that the field of complex numbers C is algebraically closed. There are many distinct and completely different proofs of this result. In [3] twelve proofs were given covering a wide area of mathematics. In this section we use Galois theory to present a proof. Before doing this we briefly mention some of the history surrounding this theorem. The first mention of the fundamental theorem of algebra, in the form that every polynomial equation of degree n has exactly n roots, was given by Peter Roth of Nurnberg in 1608. However its conjecture is generally credited to Girard who also stated the result in 1629. It was then more clearly stated by Descartes in 1637 who also distinguished between real and imaginary roots. The first published proof of the fundamental theorem of algebra was then given by D’Alembert in 1746. However there were gaps in D’Alembert’s proof and the first fully accepted proof was that given by Gauss in 1797 in his Ph.D. thesis. This was published in 1799. Interestingly enough, in reviewing Gauss’ original proof, modern scholars tend to agree that there are as many holes in this proof as in D’Alembert’s proof. Gauss, however, published three other proofs with no such holes. He published second and third proofs in 1816 while his final proof, which was essentially another version of the first, was presented in 1849. Theorem 17.7.1. Each nonconstant polynomial f .x/ 2 CŒx, where C is the field of complex numbers, has a zero in C. Therefore C is an algebraically closed field.

262

Chapter 17 Applications of Galois Theory

Proof. Let f .x/ 2 CŒx be a nonconstant polynomial and let K be the splitting field of f .x/ over C. Since the characteristic of the complex numbers C is zero this will be a Galois extension of C. Since C is a finite extension of R this field K would also be a Galois extension of R. The fundamental theorem of algebra asserts that K must be C itself, and hence the fundamental theorem of algebra is equivalent to the fact that any nontrivial Galois extension of C must be C. Let K be any finite extension of R with jK W Rj D 2m q; .2; q/ D 1. If m D 0, then K is an odd-degree extension of R. Since K is separable over R, from the primitive element theorem it is a simple extension, and hence K D R.˛/, where the minimal polynomial m˛ .x/ over R has odd degree. However, odd-degree real polynomials always have a real root, and therefore m˛ .x/ is irreducible only if its degree is one. But then ˛ 2 R and K D R. Therefore, if K is a nontrivial finite extension of R of degree 2m q we must have m > 0. This shows more generally that there are no odd-degree finite extensions of R. Suppose that K is a degree 2 extension of C. Then K D C.˛/ with deg m˛ .x/ D 2 where m˛ .x/ is the minimal polynomial of ˛ over C. But from the quadratic formula complex quadratic polynomials always have roots in C so a contradiction. Therefore, C has no degree 2 extensions. Now, let K be a Galois extension of C. Then K is also Galois over R. Suppose jK W Rj D 2m q, .2; q/ D 1. From the argument above we must have m > 0. Let G D Gal.K=R/ be the Galois group. Then jGj D 2m q, m > 0, .2; q/ D 1. Thus G has a 2-Sylow subgroup of order 2m and index q (see Theorem 13.3.4). This would correspond to an intermediate field E with jK W Ej D 2m and jE W Rj D q. However, then E is an odd-degree finite extension of R. It follows that q D 1 and E D R. Therefore, jK W Rj D 2m and jGj D 2m . Now, jK W Cj D 2m1 and suppose G1 D Gal.K=C/. This is a 2-group. If it were not trivial, then from Theorem 13.4.1 there would exist a subgroup of order 2m2 and index 2. This would correspond to an intermediate field E of degree 2 over C. However from the argument above C has no degree 2 extensions. It follows then that G1 is trivial, that is, jG1 j D 1, so jK W Cj D 1 and K D C completing the proof. The fact that C is algebraically closed limits the possible algebraic extensions of the reals. Corollary 17.7.2. Let K be a finite field extension of the real numbers R. Then K D R or K D C. Proof. Since jK W Rj < 1 by the primitive element theorem K D R.˛/ for some ˛ 2 K. Then the minimal polynomial m˛ .x/ of ˛ over R is in RŒx and hence in CŒx. Therefore form the fundamental theorem of algebra it has a root in C. Hence ˛ 2 C. If ˛ 2 R then K D R, if not then K D C.

263

Section 17.8 Exercises

17.8

Exercises

1. For f .x/ 2 QŒx with f .x/ D x 6  12x 4 C 36x 2  50 .f .x/ D 4x 4  12x 2 C 20x  3/ 1 m

determine for each complex zero ˛ of f .x/ a finite number of radicals i D ˇi i , i D 1; : : : ; r, and a presentation of ˛ as a rational function in 1 ; : : : ; r over Q such that iC1 is irreducible over Q. 1 ; : : : ; i / and ˇi C1 2 Q. 1 ; : : : ; i / for i D 0; : : : ; r  1. 2. Let K be a field of prime characteristic p. Let n 2 N and Kn the splitting field of x n  1 over K. Show that Aut.Kn jK/ is cyclic. 3. Let f .x/ D x 4  x C 1 2 ZŒx. Show: (i) f has a real zero. (ii) f is irreducible over Q. (iii) If u C iv (u; v 2 R) is a zero of f in C, then g D x 3  4x  1 is the minimal polynomial of 4u2 over Q. (iv) The Galois group of f over Q has an element of order 3. (v) No zero a 2 C of f is constructible from the points 0 and 1 with straightedge and compass. 4. Show that each polynomial f .x/ over R decomposes in linear factors and quadratic factors (f .x/ D d.x  a1 /  .x  a2 /    .x 2 C b1 x C c1 /  .x 2 C b2 x C c2 /    , d 2 R). 5. Let E be a finite (commutative) field extension of R. Then E Š R or E Š C. 6. Let n  1 be a natural number and x an indeterminate over C. Consider the polynomial x n  1 2 ZŒx. In CŒx it decomposes in linear factors: x n  1 D .x  1 /.x  2 /    .x  n /; where the complex numbers 

 D e 2 i n D cos

2 2 C i  sin ; n n

1    n;

are all (different) n-th roots of unity, that is especially n D 1. These  form a from 1 generated multiplicative cyclic group G D ¹1 ; 2 ; : : : ; n º. It is  D 1 . An n-th root of unity  is called a primitive n-th root of unity, if  is not an m-th root of unity for any m < n.

264

Chapter 17 Applications of Galois Theory

Show that the following are equivalent: (i)  is a primitive n-th root of unity. (ii)  is a generating element of G. (iii) gcd.; n/=1. 7. The polynomial n .x/ 2 CŒx, whose zeros are exactly the primitive n-th roots of unity, is called the n-th cyclotomic polynomial. With Exercise 6 it is: Y Y  .x   / D .x  e 2 i n /: n .x/ D 1n gcd.;n/D1

1n gcd.;n/D1

The degree of n .x/ is the number of the integers ¹1; : : : ; nº, which are coprime to n. Show: Q (i) x n  1 D d 1 d .x/. d jn

(ii) n .x/ 2 ZŒx for all n  1. (iii) n .x/ is irreducible over Q (and therefore also over Z) for all n  1. 8. Show that the Fermat numbers F0 ; F1 ; F2 ; F3 ; F4 are all prime but F5 is composite and divisible by 641.

Chapter 18

The Theory of Modules

18.1

Modules Over Rings

Recall that a vector space V over a field F is an abelian group V with a scalar multiplication  W F  V ! V satisfying (1) f .v1 C v2 / D f v1 C f v2 for f 2 F and v1 ; v2 2 V . (2) .f1 C f2 /v D f1 v C f2 v for f1 ; f2 2 F and v 2 V . (3) .f1 f2 /v D f1 .f2 v/ for f1 ; f2 2 F and v 2 V . (4) 1v D v for v 2 V . Vector spaces are the fundamental algebraic structures in linear algebra and the study of linear equations. Vector spaces have been crucial in our study of fields and Galois theory since any field extension is a vector space over any subfield. In this context the degree of a field extension is just the dimension of the extension field as a vector space over the base field. If we modify the definition of a vector space to allow scalar multiplication from an arbitrary ring we obtain a more general structure called a module. We will formally define this below. Modules generalize vector spaces but the fact that the scalars do not necessarily have inverses makes the study of modules much more complicated. Modules will play an important role in both the study of rings and the study of abelian groups. In fact any abelian group is a module over the integers Z so that modules, besides being generalizations of vector spaces can also be considered as generalizations of abelian groups. In this chapter we will introduce the theory of modules. In particular we will extend to modules the basic algebraic properties such as the isomorphism theorems that have been introduced earlier for groups, rings and fields. In this chapter we restrict ourselves to commutative rings so that throughout R is always a commutative ring. If R has an identity 1 then we always consider only the case that 1 ¤ 0. Throughout this chapter we use letters a; b; c; m; : : : for ideals in R. For principal ideals we write hai or aR for the ideal generated by a 2 R. We note however that the definition can be extended to include modules over noncommutative rings. In this case we would speak of left modules and right modules. Definition 18.1.1. Let R D .R; C;  / a commutative ring and M D .M; C/ an abelian group. M together with a scalar multiplication  W RM ! M; .˛; x/ 7! ˛x, is called a R-module or module over R if the following axioms hold:

266

Chapter 18 The Theory of Modules

(M1) .˛ C ˇ/x D ˛x C ˇx, (M2) ˛.x C y/ D ˛x C ˛y and (M3) .˛ˇ/x D ˛.ˇx/ for all ˛; ˇ 2 R and x; y 2 M . If R has an identity 1 then M is called an unitary R-module if in addition (M4) 1  x D x for all x 2 M holds. In the following, R always is a commutative ring. If R contains an identity 1 then M always is an unitary R-module. If R has an identity 1 then we always assume 1 ¤ 0. As usual we have the rules: 0  x D 0;

˛  0 D 0;

.˛x/ D .˛/x D ˛.x/

for all ˛ 2 R and for all x 2 M . We next present a series of examples of modules. Example 18.1.2.

(1) If R D K is a field then a K-module is a K-vector space.

(2) Let G D .G; C/ be an abelian group. If n 2 Z and x 2 G then nx is defined as usual: 0  x D 0; nx D x  C… x „ C ƒ‚

if n > 0

and

n-times

nx D .n/.x/

if n < 0:

Then G is an unitary Z-module via the scalar multiplication  W Z  G ! G;

.n; x/ 7! nx:

(3) Let S be a subring of R. Then via .s; r/ 7! sr the ring R itself becomes an S-module. (4) Let KPbe a field, V a K-vector space and P f W Vi ! V a linear map of V . Let i p D i ˛i t 2 KŒt . Then p.f / WD i ˛i f defines a linear map of V and V is an unitary KŒt -module via the scalar multiplication KŒt   V ! V;

.p; v/ 7! pv WD p.f /.v/:

(5) If R is a commutative ring and a is an ideal in R then a is a module over R. Basic to all algebraic theory is the concept of substructures. Next we define submodules.

Section 18.1 Modules Over Rings

267

Definition 18.1.3. Let M be an R-module. ; ¤ U  M is called a submodule of M if (UMI) .U; C/ < .M; C/ and (UMII) ˛ 2 R; u 2 U ) ˛u 2 U , that is, RU  U . Example 18.1.4. (1) In an abelian group G, considered as a Z-module, the subgroups are precisely the submodules. (2) The submodules of R, considered as a R-module, are precisely the ideals. (3) Rx WD ¹˛x W ˛ 2 Rº is a submodule of M for each x 2 M . (4) Let K be a field, V a K-vector space and f W V ! V a linear map of V . Let U be a submodule of V , considered as a KŒt -module as above. Then the following holds: (a) U < V . (b) pU D p.f /U  U for all p 2 KŒt . Especially ˛U  U for p D ˛ 2 K and t U D f .U /  U for p D t , that is, U is an f -invariant subspace. On the other side also, p.f /U  U for all p 2 KŒt  if U is an f -invariant subspace. We next extend to modules the concept of a generating system. For a single generator, as with groups, this is called cyclic. Definition 18.1.5. A submodule U of the R-module M is called cyclic if there exists an x 2 M with U D Rx. As in vector spaces, groups and rings the following constructions are standard leading us to generating systems. T (1) Let M be a R-module and ¹Ui W i 2 I º a family of submodules. Then i 2I Ui is a submodule of M . (2) Let M be a R-module. If A  M then we define hAi WD

\

¹U W U submodule of M with A  U º:

hAi is the smallest submodule of M which contains P A. If R has an identity 1 then hAi is the set of all linear combinations i ˛i ai with all ˛i 2 R, all ai 2 A. This holds because M is unitary and na D n.1  a/ D .n  1/a for n 2 Z and a 2 A, that is, we may consider the pseudoproduct na as a real product in the module. Especially, if R has a unit 1 then aR D h¹aºi DW hai.

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Chapter 18 The Theory of Modules

Definition 18.1.6. Let R have an identity 1. If M D hAi then A is called a generating system of M . M is called finitely generated if there are a1 ; : : : ; an 2 M with M D h¹a1 ; : : : ; an ºi DW ha1 ; : : : ; an i. The following is clear. Lemma 18.1.7. Let Ui be submodules of M , i 2 I , I an index set. Then [ ²X ³ Ui D ai W ai 2 Ui ; L  I finite : i 2I

S

i 2L

P

the Ui . A sum PWe write h i 2I Ui i DW i 2I Ui and call this submodule the sum ofP ai , ai 2 Ui , i2I Ui is called a direct sum if for each representation Pof 0 as 0 D it follows L that all ai D 0. This is equivalent to Ui \ i ¤j Uj D 0 for all i 2 I . Notation: i 2I Ui ; and if I D ¹1; : : : ; nº then we write U1 ˚    ˚ Un , too. In analogy with our previously defined algebraic structure we extend to modules the concepts of quotient modules and module homomorphisms. Definition 18.1.8. Let U be a submodule of the R-module M . Let M=U be the factor group. We define a (well-defined) scalar multiplication R  M=U ! M=U;

˛.x C U / WD ˛x C U:

With this M=U is a R-module, the factor module or quotient module of M by U . In M=U we have the operations .x C U / C .y C U / D .x C y/ C U and ˛.x C U / D ˛x C U: A module M over a ring R can also be considered as a module over a quotient ring of R. The following is straightforward to verify (see exercises). Lemma 18.1.9. Let P a C R an ideal in R and M a R-module. The set of all finite sums of the form ˛i xi , ˛i 2 a, xi 2 M , is a submodule of M which we denote by aM . The factor group M=aM becomes a R=a-module via the well-defined scalar multiplication .˛ C a/.m C aM / D ˛m C aM: If here R has an identity 1 and a is a maximal ideal then M=aM becomes a vector space over the field K D R=a.

269

Section 18.1 Modules Over Rings

We next define module homomorphisms Definition 18.1.10. Let R be a ring and M , N be R-modules. A map f W M ! N is called a R-module homomorphism (or R-linear) if f .x C y/ D f .x/ C f .y/ and f .˛x/ D ˛f .x/ for all ˛ 2 R and all x; y 2 M . Endo-, epi-, mono-, iso- and automorphisms are defined analogously via the corresponding properties of the maps. If f W M ! N and g W N ! P are module homomorphisms then g ı f W M ! P is also a module homomorphism. If f W M ! N is an isomorphism then also f 1 W N ! M . We define kernel and image in the usual way: ker.f / WD ¹x 2 M W f .x/ D 0º and im.f / WD f .M / D ¹f .x/ W x 2 M º: ker.f / is a submodule of M and im.f / is a submodule of N . As usual: f is injective ” ker.f / D ¹0º: If U is a submodule of M then the map x 7! x C U defines a module epimorphism (the canonical epimorphism) from M onto M=U with kernel U . There are module isomorphism theorems. The proofs are straightforward extensions of the corresponding proofs for groups and rings. Theorem 18.1.11 (module isomorphism theorems). Let M; N be R-modules. (1) If f W M ! N is a module homomorphism then f .M / Š M= ker.f /: (2) If U; V are submodules of the R-module M then U=.U \ V / Š .U C V /=V: (3) If U and V are submodules of the R-module M with U  V  M then .M=U /=.V =U / Š M=V:

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Chapter 18 The Theory of Modules

For the proofs, as for groups, just consider the map f W U C V ! U=.U \ V /, u C v 7! u C .U \ V / which is well-defined because U \ V is a submodule of U ; we have ker.f / D V . Note that ˛ 7! ˛ , 2 R fixed, defines a module homomorphism R ! R if we consider R itself as a R-module.

18.2

Annihilators and Torsion

In this section we define torsion for an R-module and a very important subring of R called the annihilator. Definition 18.2.1. Let M be an R-module. For a fixed a 2 M consider the map a W R ! M , a .˛/ WD ˛a. a is a module homomorphism considering R as an R-module. We call ker.a / the annihilator of a denoted Ann.a/, that is Ann.a/ D ¹˛ 2 R W ˛a D 0º: Lemma 18.2.2. Ann.a/ is a submodule of R and the module isomorphism theorem .1/ gives R= Ann.a/ Š Ra. We next extend the annihilator to whole submodules of M . Definition 18.2.3. Let U be a submodule of the R-module M . The annihilator Ann.U / is defined to be Ann.U / WD ¹˛ 2 R W ˛u D 0 for all u 2 U º: T As for single elements, since Ann.U / D u2U Ann.u/, then Ann.U / is a submodule of R. If 2 R, u 2 U , then u 2 U , that means, if u 2 Ann.U / then also u 2 Ann.U / because .˛ /u D ˛. u/ D 0. Hence, Ann.U / is an ideal in R. Suppose that G is an abelian group. Then as mentioned G is a Z-module. An element g 2 G is a torsion element or has finite order if ng D 0 for some n 2 N. The set Tor.G/ consists of all the torsion elements in G. An abelian group is torsion-free if Tor.G/ D ¹0º. Lemma 18.2.4. Let G be an abelian group. Then Tor.G/ is a subgroup of G and G= Tor.G/ is torsion-free. We extend this concept now to general modules.

Section 18.3 Direct Products and Direct Sums of Modules

271

Definition 18.2.5. The R-module M is called faithful if Ann.M / D ¹0º. An element a 2 M is called a torsion element, or element of finite order, if Ann.a/ ¤ ¹0º. A module without torsion elements ¤ 0 is called torsion-free. If the R-module M is torsion-free then R has no zero divisors ¤ 0. Theorem 18.2.6. Let R be an integral domain and M an R-module (by our agreement M is unitary). Let Tor.M / D T .M / be the set of torsion elements of M . Then Tor.M / is a submodule of M and M= Tor.M / is torsion-free. Proof. If m 2 Tor.M /, ˛ 2 Ann.m/, ˛ ¤ 0 and ˇ 2 R then we get ˛.ˇm/ D .˛ˇ/m D .ˇ˛/m D ˇ.˛m/ D 0, that is, ˇm 2 Tor.M /, because ˛ˇ ¤ 0 if ˇ ¤ 0 (R is an integral domain). Let m0 another element of Tor.M / and 0 ¤ ˛ 0 2 Ann.m0 /. Then ˛˛ 0 ¤ 0 and ˛˛ 0 .m C m0 / D ˛˛ 0 m C ˛˛ 0 m0 D ˛ 0 .˛m/ C ˛.˛ 0 m0 / D 0, that is, m C m0 2 Tor.M /. Therefore Tor.M / is a submodule. Now, let m C Tor.M / be a torsion element in M= Tor.M /. Let ˛ 2 R, ˛ ¤ 0 with ˛.m C Tor.M // D ˛m C Tor.M / D Tor.M /. Then ˛m 2 Tor.M /. Hence there exists a ˇ 2 R, ˇ ¤ 0, with 0 D ˇ.˛m/ D .ˇ˛/m. Since ˇ˛ ¤ 0 we get that m 2 Tor.M / and the torsion element m C Tor.M / is trivial.

18.3

Direct Products and Direct Sums of Modules

Let Mi , i 2 I , I ¤ ;, be a family of R-modules. On the direct product ² ³ [ Y Mi D f W I ! Mi W f .i/ 2 Mi for all i 2 I P D i 2I

i 2I

we define the module operations CWP P !P

and

 WRP !P

via .f C g/.i/ WD f .i / C g.i/ and .˛f /.i/ WD ˛f .i/: Q Together with this operations P D i 2I Mi is an R-module, the direct product of the Mi . If we identify f with the I -tuple of the images f D .fi /i 2I then the sum and the scalar multiplication are componentwise. If I D ¹1; : : : ; nº and Mi D M for Q n D all i 2 I then we write, as usual,QM i 2I Mi . M WD ¹0º. We make the agreement that i i 2I D; L Q M WD ¹f 2 M W f .i / D 0 for almost all i º (“for almost all i ” i i i2I i 2I means that there are at most finitely many i with f .i/ ¤ 0) is a submodule of the If I D ¹1; : : : ; nº then we write direct product, called the direct sumQof the Mi . L L n n M D M ˚    ˚ M . Here M D i 1 n i i2I i D1 i D1 Mi for finite I .

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Chapter 18 The Theory of Modules

Theorem 18.3.1.

(1) If  2 Per.I / is a permutation of I then Y

Y

Mi Š

i 2I

and

M

M.i/

i 2I

Mi Š

i 2I

M

M.i/ :

i 2I

S (2) If I D P j 2J Ij , the disjoint union, then Y

Mi Š

i 2I

and

M i 2I

YY j 2J

Mi Š

Mi

i 2Ij

M M j 2J



 Mi :

i 2Ij

Proof. (1) Consider the map fS7! f ı . Q (2) Consider the map f 7! j 2J fj where fj 2 i 2Ij Mi is the restriction of f S S onto Ij , and j 2J fj is on J defined by . j 2J fj /.k/ WD fk D f .k/. Q Let I ¤ ;. If M D i 2I Mi then we get in a natural manner module homomorphisms i W M ! Mi via f 7! f .i /I i is called the projectionLonto the i-thQcomponent. In duality we define module homomorphisms ıi W Mi ! i 2I Mi  i 2I Mi via ıi .mi / D .nj /j 2I where nj D 0 if i ¤ j and ni D mi . ıi is called the i-th canonical injection. If I D ¹1; : : : ; nº then i .a1 ; : : : ; ai ; : : : ; an / D ai and ıi .mi / D .0; : : : ; 0; mi ; 0; : : : ; 0/. Theorem 18.3.2 (universal properties). Let A; Mi ; i 2 I ¤ ;, be R-modules. (1) If i W A ! Mi , i 2 I , are module homomorphisms then there exists exactly one Q module homomorphism  W A ! i 2I Mi such that for each i the following diagram commutes:

that is, j D j ı  where j is the j -th projection.

Section 18.4 Free Modules

273

(2) If ‰i W Mi ! A, i 2 I , are module L homomorphisms then there exists exactly one module homomorphism ‰ W i 2I Mi ! A such that for each j 2 J the following diagram commutes:

that is, ‰j D ‰ ı ıj where ıj is the j -th canonical injection. Proof. (1) If there is such  then the j Q -th component of .a/ is equal j .a/ because j ı  D j . Hence, define .a/ 2 i 2I Mi via .a/.i/ WD i .a/, and  is the desired map. P // D ‰. i 2I ıi .xi // D a ‰ with ‰ ı ˛j D ‰j then ‰.x/ D ‰..xi P P(2) If there is such P i2I ‰ ı ıi .xi / D i 2I ‰i .xi /. Hence define ‰..xi // D i 2I ‰i .xi /, and ‰ is the desired map (recall that the sum is well defined).

18.4

Free Modules

If V is a vector space over a field F then V always has a basis over F which may be infinite. Despite the similarity to vector spaces, because the scalars may not have inverses this is not necessarily true for modules. We now define a basis for a module and show that only free modules have bases. LetPR be a ring with identity 1, M be a unitary R-module and S  M . Each finite sum ˛i si , the ˛i 2 R and the si 2 S, is called a linear combination in S. Since M is unitary and S ¤ ; then hSi is exactly the set of all linear combinations in S. In the following we assume that S ¤ ;. If S D ; then hS i D h;i D ¹0º, and this case is not interesting. For P convention, in the following we always assume mi ¤ mj if i ¤ j in a finite sum ˛i mi with all ˛i 2 R and all mi 2 M . : : ; mn º  M is called linear independent or Definition 18.4.1. A finite set ¹m1 ; : P free (over R) if a representation 0 D niD1 ˛i mi implies always ˛i D 0 for all i 2 ¹1; : : : ; nº, that is, 0 can be represented only trivially on ¹m1 ; : : : ; mn º. A nonempty subset S  M is called free (over R) if each finite subset of S is free. Definition 18.4.2. Let M be a R-module (as above). (1) S  M is called a basis of M if (a) M D hSi and (b) S is free (over R).

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Chapter 18 The Theory of Modules

(2) If M has a basis then M is called a free R-module. If S is a basis of M then M is called free on S or free with basis S . In this sense we can consider ¹0º as a free module with basis ;. 1. R  R D R2 , as R-module, is free with basis ¹.1; 0/; .0; 1/º. L 2. More general, let I ¤ ;. Then i 2I Ri with Ri D R for all i 2 I is free with basis ¹i W I ! R W i .j / D ıij ; i; j 2 I º where ´ 0 if i ¤ j ; ıij D 1 if i D j :

Example 18.4.3.

Especially, if I D ¹1; : : : ; nº then Rn D ¹.a1 ; : : : ; an / W ai 2 Rº is free with basis ¹i D .0; : : : ; 0; 1; 0; : : : ; 0/I 1  i  nº. „ ƒ‚ … i 1

3. Let G be an abelian group. If G, as a Z-module, is free on S  G, then G is called a free abelian group with basis S . If jS j D n < 1 then G Š Zn . Theorem 18.4.4. The R-module P M is free on S if and only if each m 2 M can be written uniquely ˛i si with ˛i 2 R, si 2 S. This is exactly the case L in the form when M D s2S Rs is the direct sum of the cyclic submodules Rs, and each Rs is module isomorphic to R. P because Proof. If S is a basis then each m 2 M can be written as mPD ˛i si P P ˇi si then .˛i  M D hSi. This representation is unique because if ˛i si D ˇi /si D 0, that is ˛i  ˇi D 0 for all i . P If, on the other side, we assume that the P 0  si that all ˛i D 0, representation is unique then we get from ˛i si D 0 D and therefore M is free on S. The rest of the theorem of P Pessentially is a rewriting the definition. P If each m 2 M can be written as m D ˛i si then P M D s2S Rs. If x 2 Rs 0 \P s2S;s¤s 0 Rs with s 0 2 S then x D ˛ 0 s 0 D si ¤s 0 ;si 2S ˛i si and 0 D ˛ 0 s 0  L si ¤s 0 ;si 2S ˛i si and therefore ˛ 0 D 0 and ˛i D 0 for all i. This gives M D s2S Rs. The cyclic modules Rs are isomorphic to R= Ann.s/, and Ann.s/ D ¹0º in the free modules. On the other side such modules are free on S. Corollary 18.4.5.

(1) M is free on S , M Š

L s2S

Rs , Rs D R for all s 2 S.

(2) If M is finitely generated and free then there exists an n 2 N0 such that M Š Rn D R  ˚ R …. „ ˚ ƒ‚ n-times

Proof. Part (1) is clear. We prove part (2). Let MPD hx1 ; : : : ; xr i and S a basis of M . representable on S as xi D si 2S ˛i si . Since the xi generate M Each xi is uniquely P P we get m D ˇi xi D i;j ˇi ˛j sj for arbitrary m 2 M , and we need only finitely many sj to generate M . Hence S is finite.

275

Section 18.4 Free Modules

Theorem 18.4.6. Let R be a commutative ring with identity 1 and M a free Rmodule. Then any two bases of M have the same cardinality. Proof. R contains a maximal ideal m, and R=m is a field (seeL Theorem 2.3.2 and 2.4.2). Then M=mM is a vector space over R=m. From M Š s2S Rs with basis L S we get mM Š s2S ms and, hence, M  M M M=mM Š Rs =mM Š .Rs=mM / Š R=m: s2S

s2S

s2S

Hence the R=m-vector space M=mM has a basis of the cardinality of S. This gives the result. Let R be a commutative ring with identity 1 and M a free R-module. The cardinality of a basis is an invariant of M , called the rank of M or dimension of M . If rank.M / D n < 1 then this means M Š Rn . Theorem 18.4.7. Each R-module is a (module-)homomorphic image of a free Rmodule. L Proof. Let M be a R-module. We consider F WD m2M Rm with Rm D R Pfor all ˛m m m 2 M . F is a free R-module. The map f W F ! M , f ..˛m /m2M / D defines a surjective module homomorphism. Theorem 18.4.8. Let F; M be R-modules, and let F be free. Let f W M ! F be a module epimorphism. Then there exists a module homomorphism g W F ! M with f ı g D idF , and we have M D ker.f / ˚ g.F /. Proof. Let S be a basis of F . By the axiom of choice there exists for each s 2 S an element ms 2 M with f .ms / P D s (f is surjective). We define the map g W F ! M P via s 7! ms linearly, that is, g. si 2S ˛i si / D si 2S ˛i msi . Since F is free, the map g is well defined. Obviously f ıg.s/ D f .ms / D s for s 2 S, that means f ıg D idF because F is free on S . For each m 2 M we have also m D gıf .m/C.mgıf .m// where g ı f .m/ D g.f .m// 2 g.F /, and since f ı g D idF the elements of the form m  g ı f .m/ are in the kernel of f . Therefore M D g.F / C ker.f /. Now let x 2 g.F / \ ker.f /. Then x D g.y/ for some y 2 F and 0 D f .x/ D f ı g.y/ D y, and hence x D 0. Therefore the sum is direct: M D g.F / ˚ ker.f /. Corollary 18.4.9. Let M be an R-module and N a submodule such that M=N is free. Then there is a submodule N 0 of M with M D N ˚ N 0 . Proof. Apply the above theorem for the canonical map  W M ! M=N with ker./ D N .

276

18.5

Chapter 18 The Theory of Modules

Modules over Principal Ideal Domains

We now specialize to the case of modules over principal ideal domains. For the remainder of this section R is always a principal ideal domain ¤ ¹0º. We now use the notation .˛/ WD ˛R, ˛ 2 R, for the principal ideal ˛R. Theorem 18.5.1. Let M be a free R-module of finite rank over the principal ideal domain R. Then each submodule U is free of finite rank, and rank.U /  rank.M /. Proof. We prove the theorem by induction on n D rank.M /. The theorem certainly holds if n D 0. Now let n  1 and assume that the theorem holds for all free Rmodules of rank < n. Let M be a free R-module of rank n with basis ¹x1 ; : : : ; xn º. Let U be a submodule of M . We represent the elements of U as linear combination of the basis elements x1 ; : : : ; xn , and we consider the set of coefficients of x1 for the elements of U : ² ³ n X a D ˇ 2 R W ˇx1 C ˇ i xi 2 U : i D2

Certainly a is an ideal in R. Since R is a principal ideal domain we have a D .˛1 / for some ˛1 2 R. Let u 2 U be an element in U which has ˛1 as its first coefficient, that is n X ˛i xi 2 U: u D ˛1 x1 C i D2

Let v 2 U be arbitrary. Then v D .˛1 x1 / C

n X

i xi :

i D2

Hence v  u 2 U 0 WD U \M 0 where M 0 is the free R-module with basis ¹x2 ; : : : ; xn º. By induction, U 0 is a free submodule of M 0 with a basis ¹y1 ; : : : ; y t º, t  n  1. If ˛1 D 0 then a D .0/ and U D U 0 , and there is nothing to prove. Now let ˛1 ¤ 0. We show that ¹u; y1 ; : : : ; y t º is a basisPof U . v  u is a linear combination of basis Pthe t t 0 , that is, v  u D elements of U i D1 i yi uniquely. P Hence v D u C i D1 i yi and U D hu; y1 ; : : : ; y t i. Now let be 0 D u C ti D1 i yi . We write u and the yi as linear combinations in the basis elements x1 ; : : : ; xn of M . There is only an x1 -portion in u. Hence n X 0 D ˛1 x1 C 0i xi : i D2

Therefore first ˛1 x1 D 0, that is, D 0 because R has no zero divisor ¤ 0, and further 02 D    D 0n D 0, that means, 1 D    D  t D 0.

277

Section 18.5 Modules over Principal Ideal Domains

Let R be a principal ideal domain. Then the annihilator Ann.x/ in R-modules M has certain further properties. Let x 2 M . By definition Ann.x/ D ¹˛ 2 R W ˛x D 0º G R;

an ideal in R;

hence Ann.x/ D .ıx /. If x D 0 then .ıx / D R. ıx is called the order of x and .ıx / the order ideal of x. ıx is uniquely determined up to units in R (that is, up to elements T  with 0 D 1 for some 0 2 R). For a submodule U of M we call Ann.U / D u2U .ıu / D ./ the order ideal of U . In an abelian group G, considered as a Z-module, this order for elements corresponds exactly with the order as group elements if we choose ıx  0 for x 2 G. Theorem 18.5.2. Let R be a principal ideal domain and M be a finitely generated torsion-free R-module. Then M is free. Proof. Let M D hx1 ; : : : ; xn i torsion-free and R a principal ideal domain. Each submodule hxi i D Rxi is free because M is torsion-free. We call a subset S  hx1 ; : : : ; xn i free if the submodule hSi is free. Since hxi i is free there exist such nonempty subsets. Under all free subsets S  hx1 ; : : : ; xn i we choose one with a maximal number of elements. We may assume that ¹x1 ; : : : ; xs º, 1  s  n, is such a maximal set – after possible renaming. If s D n then the theorem holds. Now, let s < n. By the choice of s the sets ¹x1 ; : : : ; xs ; xj º with s < j  n are not free. Hence there are ˛j 2 R and ˛i 2 R, not all 0, with ˛j xj D

s X

˛ i xi ;

˛j ¤ 0; s < j  n:

i D1

For the product ˛ WD ˛sC1    ˛n ¤ 0 we get ˛xj 2 Rx1 ˚    ˚ Rxs DW F , s < j  n, because ˛xi 2 F for 1  i  s. Altogether we get ˛M  F . ˛M is a submodule of the free R-module F of rank s. By Theorem 18.5.1 we have that ˛M is free. Since ˛ ¤ 0 and M is torsion-free, the map M ! ˛M , x 7! ˛x, defines an (module) isomorphism, that is, M Š ˛M . Therefore, also M is free. We remind that for an integral domain R the set Tor.M / D T .M / D ¹x 2 M W 9˛ 2 R; ˛ ¤ 0; with ˛x D 0º of the torsion elements of an R-module M is a submodule with torsion-free factor module M=T .M /. Corollary 18.5.3. Let R be a principal ideal domain and M be a finitely generated R-module. Then M D T .M / ˚ F with a free submodule F Š M=T .M /. Proof. M=T .M / is a finitely generated, torsion-free R-module, and, hence, free. By Corollary 18.4.9 we have M D T .M / ˚ F , F Š M=T .M /.

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Chapter 18 The Theory of Modules

From now on we are interested in the case that M ¤ ¹0º is a torsion R-module, that is, M D T .M /. Let R be a principal ideal domain and M D T .M / an Rmodule. Let M ¤ ¹0º and finitely generated. As above, let ıx the order of x 2 M – unique up T to units in R – and .ıx / D ¹˛ 2 R W ˛x D 0º the order ideal of x. Let ./ D x2M .ıx / the order ideal of M . Since ./  .ıx / we have ıx j for all x 2 M . Since principal ideal domains are unique factorization domains, if  ¤ 0 then there can not be many essentially different orders (that means different up to units). Since M ¤ ¹0º and finitely generated we have in any case  ¤ 0, because if M D hx1 ; : : : ; xn i, ˛i xi D 0 with ˛i ¤ 0 then ˛M D ¹0º if ˛ WD ˛1    ˛n ¤ 0. Lemma 18.5.4. Let R be a principal ideal domain and M ¤ ¹0º be an R-module with M D T .M /. (1) If the orders ıx and ıy of x; y 2 M are relatively prime, that is, gcd.ıx ; ıy / D 1, then .ıxCy / D .ıx ıy /. (2) Let ız be the order of z 2 M , z ¤ 0. If ız D ˛ˇ with gcd.˛; ˇ/ D 1 then there exist x; y 2 M with z D x C y and .ıx / D .˛/, .ıy / D .ˇ/. Proof. (1) Since ıx ıy .x C y/ D ıx ıy x C ıx ıy y D ıy ıx x C ıx ıy y D 0 we get .ıx ıy /  .ıxCy /. On the other side, from ıx x D 0 and ıxCy .x C y/ D 0 we get 0 D ıx ıxCy .x C y/ D ıx ıxCy y, that means, ıx ıxCy 2 .ıy / and, hence, ıy jıx ıxCy . Since gcd.ıx ; ıy / D 1 we have ıy jıxCy . Analogously ıx jıxCy . Hence, ıx ıy jıxCy and .ıxCy /  .ıx ıy /. (2) Let ız D ˛ˇ with gcd.˛; ˇ/ D 1. Then there are ; 2 R with 1 D ˛ C ˇ. Therefore we get z D 1  z D ˛z C ˇz D y C x D x C y: „ƒ‚… „ƒ‚… DWy

DWx

Since ˛x D ˛ ˇz D ız z D 0 we get ˛ 2 .ız /, that means, ıx j˛. On the other side, from 0 D ıx x D ˇıx z we get ız j ˇıx and hence ˛ˇj ˇıx because ız D ˛ˇ. Therefore ˛j ıx . From gcd.˛; / D 1 we get ˛jıx . Therefore ˛ is associated to ıx , that is ˛ D ıx  with  a unit in R and further .˛/ D .ıx /. Analogously .ˇ/ D .ıy /. In Lemma 18.5.4 we do not need M D T .M /. We only need x; y; z 2 M with ıx ¤ 0, ıy ¤ 0 and ız ¤ 0, respectively. Corollary 18.5.5. Let R be a principal ideal domain and M ¤ ¹0º be an R-module with M D T .M /. 1. Let x1 ; : : : ; xn 2 M be pairwise different and pairwise relatively prime orders ıxi D ˛i . Then y D x1 C    C xn has order ˛ WD ˛1 : : : ˛n .

Section 18.6 The Fundamental Theorem for Finitely Generated Modules

279

2. Let 0 ¤ x 2 M and ıx D 1k1 : : : nkn be a prime decomposition of the order ıx of x ( a unit in R and the i pairwise nonassociate prime elements in R) where n > 0, ki > 0. Then there exist xi , i D 1; : : : ; n, with ıxi associated to iki and x D x1 C    C xn .

18.6

The Fundamental Theorem for Finitely Generated Modules

In Section 10.4 we described the following result called the basis theorem for finite abelian groups (in the following we give a complete proof in detail; an elementary proof is given in Chapter 19.). Theorem 18.6.1 (Theorem 10.4.1, basis theorem for finite abelian groups). Let G be a finite abelian group. Then G is a direct product of cyclic groups of prime power order. This allowed us, for a given finite order n, to present a complete classification of abelian groups of order n. In this section we extend this result to general modules over principal ideal domains. As a consequence we obtain the fundamental decomposition theorem for finitely generated (not necessarily finite) abelian groups, which finally proves Theorem 10.4.1. In the next chapter we present a separate proof of this in a slightly different format. Definition 18.6.2. Let R be a principal ideal domain and M be an R-module. Let  2 R be a prime element. M WD ¹x 2 M W 9k  0 with  k x D 0º is called the -primary component of M . If M D M for some prime element  2 R then M is called -primary. We certainly have the following. 1. M is a submodule of M . 2. The primary components correspond to the p-subgroup inabelian groups. Theorem 18.6.3. Let R be a principal ideal domain and M ¤ ¹0º be an R-module with M D T .M /. Then M is the direct sum of its -primary components. 1    nkn be a prime decomposition Proof. x 2 M has finite order ıx . Let ıx D 1kP of ıx . P By Corollary 18.5.5 we have that x D xi with xi 2 Mi . That means M where P is the set of the prime elements M D  2P P of R. Let y 2 M \ P k for some k  0 and y D M , that is, ı D  xi with xi 2 M i , that y 2P; ¤ l i means ıxi D for some li  0. By Corollary 18.5.5 we get that y has the order Q Q li li k i ¤ i , that means,  is associated to i ¤ i . Therefore k D li D 0 for all i, and the sum is direct.

280

Chapter 18 The Theory of Modules

If R is a principal ideal domain and ¹0º ¤ M D T .M / a finitely generated torsion R-module then there are only finitely many -primary components, that is to say for the prime elements  with j where ./ is the order ideal of M . Corollary 18.6.4. Let R be a principal ideal domain and ¹0º ¤ M be a finitely generated torsion R-module. Then M has only finitely many nontrivial primary components M1 ; : : : ; Mn , and we have M D

n M

Mi :

i D1

Hence we have a reduction of the decomposition problem to the primary components. Theorem 18.6.5. Let R be a principal ideal domain,  2 R a prime element and M ¤ ¹0º be a R-module with  k M D ¹0º; further let m 2 M with .ım / D . k /. Then there exists a submodule N  M with M D Rm ˚ N . Proof. By Zorn’s lemma the set ¹U W U submodule of M and U \ Rm D ¹0ºº has a maximal element N . This set is nonempty because it contains ¹0º. We consider M 0 WD N ˚ Rm  M and have to show that M 0 D M . Assume that M 0 ¤ M . Then there exists a x 2 M with x … M 0 , especially x … N . Then N is properly contained in the submodule Rx C N D hx; N i. By our choice of N we get A WD .Rx C N / \ Rm ¤ ¹0º. If z 2 A, z ¤ 0, then z D m D ˛x C n with , ˛ 2 R and n 2 N . Since z ¤ 0 we have m ¤ 0; also x ¤ 0 because otherwise z 2 Rm \ N D ¹0º. ˛ is not a unit in R because otherwise x D ˛ 1 . m  n/ 2 M 0 . Hence we have: If x 2 M , x … M 0 then there exist ˛ 2 R, ˛ ¤ 0, ˛ not a unit in R, 2 R with m ¤ 0 and n 2 N such that ˛x D m C n:

(?)

Especially ˛x 2 M 0 . Now let ˛ D 1 : : : r be a prime decomposition. We consider one after the other the elements x; r x; r1 r x; : : : ; 1 : : : r x D ˛x. We have x … M 0 but ˛x 2 M 0 ; hence there exists an y … M 0 with i y 2 N C Rm. 1. i ¤ ,  the prime element in the statement of the theorem. Then gcd.i ;  k / D 1, hence there are , 0 2 R with i C 0  k D 1, and we get Rm D .Ri C R k /m D i Rm because  k m D 0. Therefore i y 2 M 0 D N ˚ Rm D N C i Rm. 2. i D . Then we write y as y D n C m with n 2 N and  2 R. This is possible because y 2 M 0 . Since  k M D ¹0º we get 0 D  k1  y D  k1 nC k1 m. Therefore  k1 n D  k1 m D 0 because N \Rm D ¹0º. Especially we get  k1  2 .ım /, that is,  k j k1  and, hence, j. Therefore y D n C m D n C 0 m 2 N C Rm, 0 2 R.

Section 18.6 The Fundamental Theorem for Finitely Generated Modules

281

Hence, in any case we have i y 2 N C i Rm, that is, i y D n C i z with n 2 N and z 2 Rm. It follows i .y  z/ D n 2 N . y  z is not an element of M 0 because y … M 0 . By (?) we have therefore ˛; ˇ 2 R, ˇ ¤ 0 not a unit in R with ˇ.y  z/ D n0 C ˛m, ˛m ¤ 0, n0 2 N . We write z 0 D ˛m, then z 0 2 Rm, z 0 ¤ 0, and ˇ.y  z/ D n0 C z 0 . So, we have the equations ˇ.y  z/ D n0 C z 0 , z 0 ¤ 0, and i .y  z/ D n:

(??)

We have gcd.ˇ; i / D 1 because otherwise i jˇ and, hence, ˇ.y  z/ 2 N and z 0 D 0 because N \ Rm D ¹0º. Then there exist ; 0 with i C 0 ˇ D 1. In (??) we multiply the first equation with 0 and the second with . Addition gives y  z 2 N ˚ Rm D M 0 , and hence y 2 M 0 which contradicts y … M 0 . Therefore M D M 0 . Theorem 18.6.6. Let R be a principal ideal domain,  2 R a prime element and M ¤ ¹0º a finitely generated L-primary R-module. Then there exist finitely many m1 ; : : : ; ms 2 M with M D siD1 Rmi . Proof. Let M D hx1 ; : : : ; xn i. Each xi has an order  ki . We may assume that k1 D max¹k1 ; k2 ; : : : ; kn º, possibly after renaming. We have  ki xi D 0 for all i. Since xik1 D .xiki /k1 ki we have also  k1 M D 0, and also .ıx1 / D . k1 /. Then M D Rx1 ˚ N for some submodule N  M by Theorem 18.6.5.Now N Š M=Rx1 and M=Rx1 is generated by the elements x2 C Rx1 ; : : : ; xn C Rx1 . Hence, N is finitely generated by n  1 elements; and certainly N is -primary. This proves the result by induction. Since Rmi Š R= Ann.mi / and Ann.mi / D .ımi / D . ki / we get the following extension of Theorem 18.6.6. Theorem 18.6.7. Let R be a principal ideal domain,  2 R a prime element and M ¤ ¹0º a finitely generated -primary R-module. Then there exist finitely many k1 ; : : : ; ks 2 N with s M R=. ki /; M Š i D0

and M is, up to isomorphism, uniquely determined by .k1 ; : : : ; ks /. L Proof. The first part, that is, a description as M Š siD0 R=. ki / follows directly from Theorem 18.6.6. Now, let M Š

n M i D0

ki

R=. / Š

m M i D0

R=. li /:

282

Chapter 18 The Theory of Modules

We may assume that k1  k2      kn > 0 and l1  l2  L    lm > 0. We n ki consider first the submodule N WD ¹x 2 M W x D 0º. Let M D i D1 R=. /. If P k k 1 we writeL then x D .ri C . i //Lwe have x D 0 if and only if ri 2 . i /, that is, N Š niD1 . ki 1 /=. ki / Š niD1 R=./ because  k1 R= k R Š R=R. Since .˛ C .//x D ˛x if x D 0 we get that N is an R=./-module, and, hence, a vector space over the field R=./. From the decompositions N Š

n M

R=./

and, analogously,

N Š

i D1

m M

R=./

i D1

we get n D dimR=./ N D m:

(???)

Assume that there is an i with ki < li or li < ki . Without loss of generality assume that there is an i with ki < li . Let j be the smallest index for which kj < lj . Then (because of the ordering of the ki ) 0

kj

M WD  M Š

n M i D1

kj

ki

 R= R Š

j 1 M

 kj R= ki R;

i D1

because if i > j then  kj R= ki R D ¹0º. We now M 0 D  kj M with respect to the second decomposition, that is, Lm consider 0 k j M Š iD1  R= li R. By our choice of j we have kj < lj  li for 1  i  j . Therefore, in this second decomposition, the first j summands  kj R= li R are unequal ¹0º, that is  kj R= li R ¤ ¹0º if 1  i  j . The remaining summands are ¹0º or of the form R= s R. Hence, altogether, on the one side M 0 is a direct sum of j  1 cyclic submodules and on the other side a direct sum of t  j nontrivial submodules. But this contradicts the above result (???) about the number of direct sums for finitely generated -primary modules because, certainly, M 0 is also finitely generated and -primary. Therefore ki D li for i D 1; : : : ; n. This proves the theorem. Theorem 18.6.8 (fundamental theorem for finitely generated modules over principal ideal domains). Let R be a principal ideal domain and M ¤ ¹0º be a finitely generated (unitary) R-module. Then there exist prime elements 1 ; : : : ; r 2 R, 0  r < 1 and numbers k1 ; : : : ; kr 2 t N, t 2 N0 such that  ˚ R M Š R=.1k1 / ˚ R=.2k2 / ˚    ˚ R=.rkr / ˚ R „ ˚ ƒ‚ …; t-times

and M is, up to isomorphism, uniquely determined by .1k1 ; : : : ; rkr ; t /.

283

Section 18.7 Exercises

The prime elements i are not necessarily pairwise different (up to units in R), that means it can be i D j for i ¤ j where  is a unit in R. Proof. The proof is a combination of the preceding results. The free part of M is isomorphic to M=T .M /, and the rank of M=T .M /, which we call here t , is uniquely determined because two bases of M=T .M / have the same cardinality. Therefore we may restrict ourselves on torsion modules. Here we have a reduction to -primary L L modules because in a decomposition M D i R=.iki / is M D i D R=.iki / the -primary component of M (an isomorphism certainly maps a -primary component onto a -primary component). So it is only necessary, now, to consider primary modules M . The uniqueness statement now follows from Theorem 18.6.8. Since abelian groups can be considered as Z-modules, and Z is a principal ideal domain, we get the following corollary. We will restate this result in the next chapter and prove a different version of it. Theorem 18.6.9 (fundamental theorem for finitely generated abelian groups). Let ¹0º ¤ G D .G; C/ be a finitely generated abelian group. Then there exist prime numbers p1 ; : : : ; pr , 0  r < 1, and numbers k1 ; : : : ; kr 2 N, t 2 N0 such that  ˚ Z G Š Z=.p1k1 Z/ ˚    ˚ Z=.prkr Z/ ˚ Z …; „ ˚ ƒ‚ t-times

and G is, up to isomorphism, uniquely determined by .p1k1 ; : : : ; prkr ; t /.

18.7

Exercises

1. Let M and N be isomorphic modules over a commutative ring R. Then EndR .M / and EndR .N / are isomorphic rings. (EndR .M / is the set of all R-modules endomorphisms of M .) 2. Let R be an integral domain and M an R-module with M D Tor.M / (torsion module). Show that HomR .M; R/ D 0. (HomR .M; R/ is the set of all R-module homomorphisms from M to R.) 3. Prove the isomorphism theorems for modules (1), (2) and (3) in Theorem 18.1.11 in detail. 4. Let M; M 0 ; N be R-modules, R a commutative ring. Show: (i) HomR .M ˚ M 0 ; N / Š HomR .M; N /  HomR .M 0 ; N / (ii) HomR .N; M  M 0 / Š HomR .N; M / ˚ HomR .N; M 0 /. 5. Show that two free modules having bases whose cardinalities are equal are isomorphic.

284

Chapter 18 The Theory of Modules

6. Let M be an unitary R-module (R a commutative ring) and let ¹m1 ; : : : ; ms º be a finite subset of M . Show that the following are equivalent: (i) ¹m1 ; : : : ; ms º generates M freely. (ii) ¹m1 ; : : : ; ms º is linearly independent and generates M . (iii) Every element m 2 M is uniquely expressible in the form m D with ri 2 R.

Ps

i D1 ri mi

(iv) Each Rmi is torsion-free, and M D Rm1 ˚    ˚ Rms . 7. Let R be a principal domain and M ¤ ¹0º be an R-module with M D T .M /. (i) Let x1 ; : : : ; xn 2 M be pairwise different and pairwise relatively prime orders ıxi D ˛i . Then y D x1 C    C xn has order ˛ WD ˛1 : : : ˛n . (ii) Let 0 ¤ x 2 M and ıx D 1k1    nkn be a prime decomposition of the order ıx of x ( a unit in R and the i pairwise nonassociate prime elements in R) where n > 0, ki > 0. Then there exist xi , i D 1; : : : ; n, with ıxi associated to iki and x D x1 C    C xn .

Chapter 19

Finitely Generated Abelian Groups

19.1

Finite Abelian Groups

In Chapter 10 we described the following theorem that completely provides the structure of finite abelian groups. As we saw in Chapter 18 this result is a special case of a general result on modules over principal ideal domains. Theorem 19.1.1 (Theorem 10.4.1, basis theorem for finite abelian groups). Let G be a finite abelian group. Then G is a direct product of cyclic groups of prime power order. We review two examples that show how this theorem leads to the classification of finite abelian groups. In particular this theorem allows us, for a given finite order n, to present a complete classification of abelian groups of order n. Since all cyclic groups of order n are isomorphic to .Zn ; C/, Zn D Z=nZ, we will denote a cyclic group of order n by Zn . Example 19.1.2. Classify all abelian groups of order 60. Let G be an abelian group of order 60. From Theorem 10.4.1 G must be a direct product of cyclic groups of prime power order. Now 60 D 22  3  5 so the only primes involved are 2, 3 and 5. Hence the cyclic groups involved in the direct product decomposition of G have order either 2, 4, 3 or 5 (by Lagrange’s theorem they must be divisors of 60). Therefore G must be of the form G Š Z4  Z3  Z5 or G Š Z2  Z2  Z3  Z5 : Hence up to isomorphism there are only two abelian groups of order 60. Example 19.1.3. Classify all abelian groups of order 180. Let G be an abelian group of order 180. Now 180 D 22  32  5 so the only primes involved are 2, 3 and 5. Hence the cyclic groups involved in the direct product decomposition of G have order either 2, 4, 3, 9 or 5 (by Lagrange’s theorem they must be divisors of 180). Therefore G

286

Chapter 19 Finitely Generated Abelian Groups

must be of the form G Š Z4  Z9  Z5 G Š Z2  Z2  Z9  Z5 G Š Z4  Z3  Z3  Z5 G Š Z2  Z2  Z3  Z3  Z5 : Hence up to isomorphism there are four abelian groups of order 180. The proof of Theorem 19.1.1 involves the following lemmas. We refer back to Chapter 10 or Chapter 18 for the proofs. Notice how these lemmas mirror the results for finitely generated modules over principal ideal domains considered in the last chapter. Lemma 19.1.4. Let G be a finite abelian group and let pjjGj where p is a prime. Then all the elements of G whose orders are a power of p form a normal subgroup of G. This subgroup is called the p-primary component of G, which we will denote by Gp . Lemma 19.1.5. Let G be a finite abelian group of order n. Suppose that n D p1e1    pkek with p1 ; : : : ; pk distinct primes. Then G Š Gp1      Gpk where Gpi is the pi -primary component of G. Theorem 19.1.6 (basis theorem for finite abelian groups). Let G be a finite abelian group. Then G is a direct product of cyclic groups of prime power order.

19.2

The Fundamental Theorem: p-Primary Components

In this section we use the fundamental theorem for finitely generated modules over principal ideal domains to extend the basis theorem for finite abelian groups to the more general case of finitely generated abelian groups. In this section we consider the decomposition into p-primary components, mirroring our result in the finite case. In the next chapter we present a different form of the basis theorem with a more elementary proof.

287

Section 19.2 The Fundamental Theorem: p-Primary Components

In Chapter 18 we proved the following: Theorem 19.2.1 (fundamental theorem for finitely generated modules over principal ideal domains). Let R be a principal ideal domain and M ¤ ¹0º be a finitely generated (unitary) R-module. Then there exist prime elements 1 ; : : : ; r 2 R, 0  r < 1 and numbers k1 ; : : : ; kr 2 N, t 2 N0 such that M Š R=.1k1 / ˚ R=.2k2 / ˚    ˚ R=.rkr / ˚ R  ˚ R …; „ ˚ ƒ‚ t-times

and M is, up to isomorphism, uniquely determined by .1k1 ; : : : ; rkr ; t /. The prime elements i are not necessarily pairwise different (up to units in R), that means it can be i D j for i ¤ j where  is a unit in R. Since abelian groups can be considered as Z-modules, and Z is a principal ideal domain, we get the following corollary which is extremely important in its own right. Theorem 19.2.2 (fundamental theorem for finitely generated abelian groups). Let ¹0º ¤ G D .G; C/ be a finitely generated abelian group. Then there exist prime numbers p1 ; : : : ; pr , 0  r < 1, and numbers k1 ; : : : ; kr 2 N, t 2 N0 such that  ˚ Z G Š Z=.p1k1 Z/ ˚    ˚ Z=.prkr Z/ ˚ Z …; „ ˚ ƒ‚ t-times

and G is, up to isomorphism, uniquely determined by .p1k1 ; : : : ; prkr ; t /. Notice that the number t of infinite components is unique. This is called the rank or Betti number of the abelian group G. This number plays an important role in the study of homology and cohomology groups in topology. If G D Z  Z      Z D Zr for some r we call G a free abelian group of rank r. Notice that if an abelian group G is torsion-free then the p-primary components are just the identity. It follows that in this case G is a free abelian group of finite rank. Again using module theory it follows that subgroups of this must also be free abelian and of smaller or equal rank. Notice the distinction between free abelian groups and absolutely free groups (see Chapter 14). In the free group case a nonabelian free group of finite rank contains free subgroups of all possible countable ranks. In the free abelian case however the subgroups have smaller or equal rank. We summarize this comments. Theorem 19.2.3. Let G ¤ ¹0º be a finitely generated torsion-free abelian group. Then G is a free abelian group of finite rank r that is G Š Zr . Further if H is a subgroup of G then H is also free abelian and the rank of H is smaller or equal than the rank of G.

288

19.3

Chapter 19 Finitely Generated Abelian Groups

The Fundamental Theorem: Elementary Divisors

In this section we present the fundamental theorem of finitely generated abelian groups in a slightly different form and present an elementary proof of it. In the following G is always a finitely generated abelian group. We use the addition “C” for the binary operation, that is, C W G  G ! G;

.x; y/ 7! x C y:

We also write ng instead of g n and use 0 as the symbol for the identity element in G, that is, 0 C g D g for all g 2 G. G D hg1 ; : : : ; g t i, 0  t < 1, that is, G is (finitely) generated by g1 ; : : : ; g t , is equivalent to the fact that each g 2 G can be written in the form g D n1 g1 C n2 g2 C    C n t g t , ni 2 Z. A relation between the gi with coefficients n1 ; : : : ; n t is then each equation of the form n1 g1 C    C n t g t D 0. A relation is called nontrivial if ni ¤ 0 for at least one i. A system R of relations in G is called a system of defining relations, if each relation in G is a consequence of R. The elements g1 ; : : : ; g t are called integrally linear independent if there are no nontrivial relations between them. A finite generating system ¹g1 ; : : : ; g t º of G is called a minimal generating system if there is no generating system with t  1 elements. Certainly each finitely generated group has a minimal generating system. In the following we always assume that our finitely generated abelian group G is unequal ¹0º, that is, G is nontrivial. As above, we may consider G as a finitely generated Z-module, and in this sense, the subgroups of G are precisely the submodules. Hence, it is clear what we mean if we call G a direct product G D U1      Us of its subgroups U1 ; : : : ; Us , namely, each g 2 G can be written as g D u1 C u2 C    C us with ui 2 Ui and 



s Y

Ui \

Uj

D ¹0º:

j D1;j ¤i

To emphasize the little difference between abelian groups and Z-modules we here use the notation “direct product” instead of “direct sum”. Considered as Z-modules, for finite index sets I D ¹1; : : : ; sº we have anyway s Y i D1

Ui D

s M

Ui :

i D1

Finally we use the notation Zn instead of Z=nZ, n 2 N. In general, we use Zn to be a cyclic group of order n.

Section 19.3 The Fundamental Theorem: Elementary Divisors

289

The aim in this section is to prove the following: Theorem 19.3.1 (basis theorem for finitely generated abelian groups). Let G ¤ ¹0º be a finitely generated abelian group. Then G is a direct product G Š Zk1      Zkr  U1      Us ; r  0, s  0, of cyclic subgroups with jZki j D ki for i D 1; : : : ; r, ki jki C1 for i D 1; : : : ; r  1 and Uj Š Z for j D 1; : : : ; s. Here the numbers k1 ; : : : ; kr , r and s are uniquely determined by G, that means, are k10 ; : : : ; kr0 ; r 0 and s 0 the respective numbers for a second analogous decomposition of G then r D r 0 , k1 D k10 ; : : : ; kr D kr0 and s D s 0 . The numbers ki are called the elementary divisors of G. We can have r D 0 or s D 0 (but not both because G ¤ ¹0º). If s > 0, r D 0 then G is a free abelian group of rank s (exactly the same rank if you consider G as a free Z-module of rank s). If s D 0 then G is finite, in fact: s D 0 , G is finite. We first prove some preliminary results. Lemma P19.3.2. Let G D hg1 ; : : : ; g t i, t  2, an abelian group. Then also G D hg1 C tiD2 mi gi ; g2 ; : : : ; g t i for arbitrary m2 ; : : : ; m t 2 Z. Lemma 19.3.3. Let G be a finitely generated abelian group. Among all nontrivial relations between elements of minimal generating systems of G we choose one relation m1 g1 C    C m t g t D 0

(?)

with smallest possible positive coefficient, and let this smallest coefficient be m1 . Let n1 g1 C    C n t g t D 0

(??)

be another relation between the same generators g1 ; : : : ; g t . Then (1) m1 jn1 and (2) m1 jmi for i D 1; 2; : : : ; t . Proof. (1) Assume m1 − n1 . Then n1 D q m1 C m01 with 0 < m01 < m1 . If we multiply the relation (?) with q and subtract the resulting relation from the relation (??) then we get a relation with a coefficient m01 < m1 which contradicts the choice of m1 . Hence m1 jn1 . (2) Assume m1 − m2 . Then m2 D q m1 C m02 with 0 < m02 < m2 . ¹g1 C qg2 ; g2 ; : : : ; g t º is a minimal generating system which satisfies the relation m1 .g1 C qg2 / C m02 g2 C m3 g3 C    C m t g t D 0, and this relation has a coefficient m02 < m1 . This again contradicts the choice of m1 . Hence m1 jm2 and further m1 jmi for i D 1; : : : ; t .

290

Chapter 19 Finitely Generated Abelian Groups

Lemma 19.3.4 (invariant characterization of kr for finite abelian groups G). Let G D Zk1      Zkr and Zki finite cyclic of order ki  2, i D 1; : : : ; r, with ki jki C1 for i D 1; : : : ; r  1. Then kr is the smallest natural number n such that ng D 0 for all g 2 G. kr is called the exponent or the maximal order of G. Proof. 1. Let g 2 G arbitrary, that is, g D n1 g1 C    C nr gr with gi 2 Zki . Then ki gi D 0 for i D 1; : : : ; r by the theorem of Fermat. Since ki jkr we get kr g D n1 k1 g1 C    C nr kr gr D 0. 2. Let a 2 G with Zkr D hai. Then the order of a is kr and, hence, na ¤ 0 for all 0 < n < kr . Lemma 19.3.5 (invariant characterization of s). Let G D Zk1      Zkr  U1      Us , s > 0, where the Zki are finite cyclic groups of order ki and the Uj are infinite cyclic groups. Then, s is the maximal number of integrally linear independent elements of G; s is called the rank of G. Proof. 1. Let gi 2 Ui , gi ¤ 0, for i D 1; : : : ; s. Then the g1 ; : : : ; gs are integrally linear independent because from n1 g1 C    C ns gs D 0, the ni 2 Z, we get n1 g1 2 U1 \ .U2      Us / D ¹0º, and, hence, n1 g1 D 0, that is n1 D 0, because g1 has infinite order. Analogously we get n2 D    D ns D 0. 2. LetP g1 ; : : : ; gsC1 2 G. We look for integers x1 ; : : : ; xsC1 , not all 0, such that a relation sC1 iD1 xi gi D 0 holds. Let Zki 2 hai i, Uj D hbj i. Then we may write each gi as gi D mi1 a1 C    C mir ar C ni1 b1 C    C nis bs for i D 1; : : : ; s C 1, where mij aj 2 Zkj and nil bl 2 Ul . P Case 1: all mij aj D 0. Then sC1 i D1 xi gi D 0 is equivalent to sC1 X

xi

X s

i D1

 nij bj

D

j D1

s  sC1 X X j D1

 nij xi bj D 0:

i D1

P The system sC1 i D1 nij xi D 0, j D 1; : : : ; s, of linear equations has at least one nontrivial rational solution .x1 ; : : : ; xsC1 / because we have more unknown than equations. Multiplication with the common denominator gives a nontrivial integral solution .x1 ; : : : ; xsC1 / 2 ZsC1 . For this solution we get sC1 X

xi gi D 0:

i D1

Case 2: mij aj arbitrary. Let k ¤ 0 be a common multiple of the orders kj of the cyclic groups Zkj , j D 1; : : : ; r. Then kgi D mi1 ka1 C    C mir kar Cni1 kb1 C    C nis kbs „ ƒ‚ … „ ƒ‚ … D0

D0

Section 19.3 The Fundamental Theorem: Elementary Divisors

291

for i D 1; : : : ; s C 1. By case 1 the kg1 ; : : : ; kgsC1 are integrally linear depenP x .kg dent, that is, we have integers x1 ; : : : ; xsC1 , not all 0, with sC1 i i/ D 0 D i D1 PsC1 .x k/g , and the x k are not all 0. Hence, also g ; : : : ; g are integrally i i 1 sC1 iD1 i linear dependent. Lemma 19.3.6. Let G WD Zk1      Zkr Š Zk10      Zk 0 0 DW G 0 , the Zki ; Zkj0 r cyclic groups of orders ki ¤ 1 and kj0 ¤ 1, respectively and ki jki C1 for i D 1; : : : ; r  1 and kj0 jkj0 C1 for j D 1; : : : ; r 0  1. Then r D r 0 and k1 D k10 , k2 D k20 ; : : : ; kr D kr0 . Proof. We prove this lemma by induction on the group order jGj D jG 0 j. Certainly, Lemma 19.3.6 holds if jGj  2 because then either G D ¹0º, and here r D r 0 D 0 or G Š Z2 , and here r D r 0 D 1. Now let jGj > 2. Then especially r  1. Inductively we assume that Lemma 19.3.6 holds for all finite abelian groups of order less than jGj. By Lemma 19.3.4 the number kr is invariantly characterized, that is, from G Š G 0 follows kr D kr0 0 , that is especially, Zkr Š Zk 0 0 . Then G=Zkr Š G=Zk 0 0 , that is, r r Zk1      Zkr1 Š Zk10      Zk 0 0 . Inductively r  1 D r 0  1, that is, r D r 0 , r 1 and k1 D k10 ; : : : ; kr1 D kr0 0 1 . This proves Lemma 19.3.6. We can now present the main result, which we state again, and its proof. Theorem 19.3.7 (basis theorem for finitely generated abelian groups). Let G ¤ ¹0º be a finitely generated abelian group. Then G is a direct product G Š Zk1      Zkr  U1      Us ;

r  0; s  0;

of cyclic subgroups with jZki j D ki for i D 1; : : : ; r, ki jki C1 for i D 1; : : : ; r  1 and Uj Š Z for j D 1; : : : ; s. Here the numbers k1 ; : : : ; kr ; r and s are uniquely determined by G, that means, are k10 ; : : : ; kr0 ; r 0 and s 0 the respective numbers for a second analogous decomposition of G then r D r 0 , k1 D k10 ; : : : ; kr D kr0 and s D s0. Proof. (a) We first prove the existence of the given decomposition. Let G ¤ ¹0º be a finitely generated abelian group. Let t , 0 < t < 1, be the number of elements in a minimal generating system of G. We have to show that G is decomposable as a direct product of t cyclic groups with the given description. We prove this by induction on t . If t D 1 then the basis theorem certainly is correct. Now let t  2 and assume that the assertion holds for all abelian groups with less then t generators. Case 1: There does not exist a minimal generating system of G which satisfies a nontrivial relation. Let ¹g1 ; : : : ; g t º be an arbitrary minimal generating system for G. Let Ui D hgi i. Then all Ui are infinite cyclic and we have G D U1      U t because if, for instance, U1 \ .U2 C    C U t / ¤ ¹0º then we must have a nontrivial relation between the g1 ; : : : ; g t .

292

Chapter 19 Finitely Generated Abelian Groups

Case 2: There exist minimal generating systems of G which satisfy nontrivial relations. Among all nontrivial relations between elements of minimal generating systems of G we choose one relation m1 g 1 C    C m t g t D 0

(?)

with smallest possible positive coefficient. Without loss of generality, let m1 be this coefficient. By Lemma 19.3.3 we get m2 D q2 m1 ; : : : ; m t D q t m1 . Now, P ¹g1 C tiD2 qi gi ; gP 2 ; : : : ; g t º is a minimal generating system of G by Lemma 19.3.2. Define h1 D g1 C ti D2 qi gi , then m1 h1 D 0. If n1 h1 C n2 g2 C    C n t g t D 0 is an arbitrary relation between h1 ; g2 ; : : : ; g t then m1 jn1 by Lemma 19.3.3, and, hence n1 h1 D 0. Define H1 WD hh1 i and G 0 D hg2 ; : : : ; g t i. Then G D H1  G 0 . This we can see as follows: First, each g 2 G can be written as g D m1 h1 C m2 g2 C    C m t g t D m1 h1 C g 0 with g 0 2 G 0 . Also H1 \ G 0 D ¹0º because m1 h1 D g 0 2 G 0 implies a relation n1 h1 C n2 g2 C    C n t g t D 0 and from this we get, as above, n1 h1 D g 0 D 0. Now, inductively, G 0 D Zk2      Zkr  U1      Us with Zki a cyclic group of order ki , i D 2; : : : ; r, ki jki C1 for i D 2; : : : ; r  2, Uj Š Z for j D 1; : : : ; s, and .r  1/ C s D t  1, that is, r C s D t . Further, G D H1  G 0 where H1 is cyclic of order m1 . If r  2 then we get a nontrivial relation m1 h1 C k2 h2 D 0 „ƒ‚… „ƒ‚… D0

D0

since k2 ¤ 0. Again m1 jk2 by Lemma 19.3.3. This gives the desired decomposition. (b) We now prove the uniqueness statement. Case 1: G is finite abelian. Then the claim follows from Lemma 19.3.6 Case 2: G is arbitrary finitely generated and abelian. Let T WD ¹x 2 G W jxj < 1º, that is the set of elements of G of finite order. Since G is abelian T is a subgroup of G, the so called torsion subgroup of G. If, as above, G D Zk1      Zkr  U1      Us then T D Zk1   Zkr because an element b1 C  Cbr Cc1 C  Ccs with bi 2 Zki , cj 2 Uj has finite order if and only if all cj D 0. That means: Zk1      Zkr is, independent of the special decomposition, uniquely determined by G, and hence, also the numbers r; k1 ; : : : ; kr by Lemma 19.3.6. Finally the number s, the rank of G, is uniquely determined by Lemma 19.3.5. This proves the basis theorem for finitely generated abelian groups. As a corollary we get the fundamental theorem for finitely generated abelian groups as given in Theorem 19.2.1. Theorem 19.3.8. Let ¹0º ¤ G D .G; C/ be a finitely generated abelian group. Then there exist prime numbers p1 ; : : : ; pr , 0  r < 1, and numbers k1 ; : : : ; kr 2 N, t 2 N0 such that   Z G Š Z k1      Zp kr  Z „  ƒ‚ …; p1

r

t-times

and G is, up to isomorphism, uniquely determined by .p1k1 ; : : : ; prkr ; t /.

Section 19.3 The Fundamental Theorem: Elementary Divisors

293

Proof. For the existence we only have to show that Zmn Š Zm Zn if gcd.m; n/ D 1. For this we write Un D hm C mnZi < Zmn , Um D hn C nmZi < Zmn , and Un \ Um D ¹mnZº because gcd.m; n/ D 1. Further there are h; k 2 Z with 1 D hm C k n. Hence, l C mnZ D hlm C mnZ C kln C mnZ, and therefore Zmn D U n  Um Š Z n  Z m . For the uniqueness statement we may reduce the problem to the case jGj D p k for a prime number p and k 2 N. But here the result follows directly from Lemma 19.3.6.

From this proof we automatically get the Chinese remainder theorem for the case Zn D Z=nZ. Theorem 19.3.9 (Chinese remainder theorem). Let m1 ; : : : ; mr 2 N, r  2, with gcd.mi ; mj / D 1 for i ¤ j . Define m WD m1    mr . (1)  W Zm ! Zm1      Zmr , a C mZ 7! .a C m1 Z; : : : ; a C mr Z/, defines a ring isomorphism. (2) The restriction of  on the multiplicative group of the prime residue classes defines a group isomorphism Z?m ! Z?m1      Z?mr . (3) For given a1 ; : : : ; ar 2 Z there exists modulo m exactly one x 2 Z with x  ai .mod mi / for i D 1; : : : ; r. Recall that for k 2 N a prime residue class is defined by aCkZ with gcd.a; k/ D 1. The set of prime residue classes modulo k is certainly a multiplicative group. Proof. By Theorem 19.3.1 we get that  is an additive group isomorphism which can be extended directly to a ring isomorphism via .a C mZ/.b C mZ/ 7! .ab C m1 Z; : : : ; ab C mr Z/. The remaining statements are now obvious. Let A.n/ be the number of nonisomorphic finite abelian groups of order n D p1k1    prkr , r  1, with pairwise different prime numbers p1 ; : : : ; pr and k1 ; : : : ; kr 2 N. By Theorem 19.2.2 we have A.n/ D A.p1k1 /    A.prkr /. Hence, to calculate A.n/, we have to calculate A.p m / for a prime number p m 2 N. Again, by Theorem 19.2.2, we get G Š Zp m1      Zp mk , all mi  1, if G is abelian of order p m . If we compare the orders we get m D m1 C    C mk . We may order the mi by size. A k-tuple .m1 ; : : : ; mk / with 0 < m1  m2      mk and m1 C m2 C    C mk D m is called a partition of m. From above each abelian group of order p m gives a partition .m1 ; : : : ; mk / of m for some k with 1  k  m. On the other side, each partition .m1 ; : : : ; mk / of m gives an abelian group of order p m , namely Zpm1      Zpmk . Theorem 19.2.2 shows that different partitions give nonisomorphic groups. If we define p.m/ to be the number of partitions of m then we get the following: A.p m / D p.m/ and A.p1k1    prkr / D p.k1 /    p.kr /.

294

19.4

Chapter 19 Finitely Generated Abelian Groups

Exercises

1. Let H be a finite generated abelian group, which is the homomorphic image of a torsion-free abelian group of finite rank n. Show that H is the direct sum of  n cyclic groups. 2. Determine (up to isomorphism) all groups of order p 2 (p prime) and all abelian groups of order  15. 3. Let G be an abelian group with generating elements a1 ; : : : ; a4 and defining relations 5a1 C 4a2 C a3 C 5a4 D 0 7a1 C 6a2 C 5a3 C 11a4 D 0 2a1 C 2a2 C 10a3 C 12a4 D 0 10a1 C 8a2  4a3 C 4a4 D 0: Express G as a direct product of cyclic groups. Q 4. Let G be a finite abelian group and u D g2G g the product of all elements of G. Show: If G has an element a of order 2, then u D a, otherwise u D e. Conclude from this the theorem of Wilson: .p  1/Š  1.mod p/

for each prime p:

5. Let p be a prime and G a finite abelian p-group, that is the order of all elements of G is finite and a power of p. Show that G is cyclic, if G has exactly one subgroup of order p. Is the statement still correct, if G is not abelian?

Chapter 20

Integral and Transcendental Extensions

20.1

The Ring of Algebraic Integers

Recall that a complex number ˛ is an algebraic number if it is algebraic over the rational numbers Q. That is ˛ is a zero of a polynomial p.x/ 2 QŒx. If ˛ 2 C is not algebraic then it is a transcendental number. We will let A denote the totality of algebraic numbers within the complex numbers C, and T the set of transcendentals so that C D A [ T . The set A is the algebraic closure of Q within C. The set A of algebraic numbers forms a subfield of C (see Chapter 5) and the subset A0 D A \ R of real algebraic numbers forms a subfield of R. The field A is an algebraic extension of the rationals Q, however the degree is infinite. Since each rational is algebraic it is clear p that there are algebraic numbers. Further there are irrational algebraic numbers, 2 for example, since it is a root of the irreducible polynomial x 2  2 over Q. In Chapter 5 we proved that there are uncountably infinitely many transcendental numbers (Theorem 5.5.3). However it is very difficult to prove that any particular real or complex number is actually transcendental. In Theorem 5.5.4 we showed that the real number cD

1 X 1 10j Š

j D1

is transcendental. In this section we examine a special type of algebraic number called an algebraic integer. These are the algebraic numbers that are zeros of monic integral polynomials. The set of all such algebraic integers forms a subring of C. The proofs in this section can be found in [35]. After we do this we extend the concept of an algebraic integer to a general context and define integral ring extensions. We then consider field extensions that are nonalgebraic – transcendental field extensions. Finally we will prove that the familiar numbers e and  are transcendental. Definition 20.1.1. An algebraic integer is a complex number ˛ that is a root of a monic integral polynomial. That is, ˛ 2 C is an algebraic integer if there exists f .x/ 2 ZŒx with f .x/ D x n C bn1 x n1 C    C b0 , bi 2 Z, n  1, and f .˛/ D 0. An algebraic integer is clearly an algebraic number. The following are clear.

296

Chapter 20 Integral and Transcendental Extensions

Lemma 20.1.2. If ˛ 2 C is an algebraic integer, then all its conjugates, ˛1 ; : : : ; ˛n , over Q are also algebraic integers. Lemma 20.1.3. ˛ 2 C is an algebraic integer if and only if m˛ 2 ZŒx. To prove the converse of this lemma we need the concept of a primitive integral polynomial. This is a polynomial p.x/ 2 ZŒx such that the GCD of all its coefficients is 1. The following can be proved (see exercises): (1) If f .x/ and g.x/ are primitive then so is f .x/g.x/. (2) If f .x/ 2 ZŒx is monic then it is primitive. (3) If f .x/ 2 QŒx then there exists a rational number c such that f .x/ D cf1 .x/ with f1 .x/ primitive. Now suppose f .x/ 2 ZŒx is a monic polynomial with f .˛/ D 0. Let p.x/ D m˛ .x/. Then p.x/ divides f .x/ so f .x/ D p.x/q.x/. Let p.x/ D c1 p1 .x/ with p1 .x/ primitive and let q.x/ D c2 q1 .x/ with q1 .x/ primitive. Then f .x/ D cp1 .x/q1 .x/: Since f .x/ is monic it is primitive and hence c D 1 so f .x/ D p1 .x/q1 .x/. Since p1 .x/ and q1 .x/ are integral and their product is monic they both must be monic. Since p.x/ D c1 p1 .x/ and they are both monic it follows that c1 D 1 and hence p.x/ D p1 .x/. Therefore p.x/ D m˛ .x/ is integral. When we speak of algebraic integers we will refer to the ordinary integers as rational integers. The next lemma shows the close ties between algebraic integers and rational integers. Lemma 20.1.4. If ˛ is an algebraic integer and also rational then it is a rational integer. The following ties algebraic numbers in general to corresponding algebraic integers. Notice that if q 2 Q then there exists a rational integer n such that nq 2 Z. This result generalizes this simple idea. Theorem 20.1.5. If  is an algebraic number then there exists a rational integer r ¤ 0 such that r is an algebraic integer. We saw that the set A of all algebraic numbers is a subfield of C. In the same manner the set I of all algebraic integers forms a subring of A. First an extension of the following result on algebraic numbers. Lemma 20.1.6. Suppose ˛1 ; : : : ; ˛n are the set of conjugates over Q of an algebraic integer ˛. Then any integral symmetric function of ˛1 ; : : : ; ˛n is a rational integer.

Section 20.1 The Ring of Algebraic Integers

297

Theorem 20.1.7. The set I of all algebraic integers forms a subring of A. We note that A, the field of algebraic numbers, is precisely the quotient field of the ring of algebraic integers. An algebraic number field is a finite extension of Q within C. Since any finite extension of Q is a simple extension each algebraic number field has the form K D Q. / for some algebraic number . Let K D Q. / be an algebraic number field and let RK D K \ I. Then RK forms a subring of K called the algebraic integers or just integers of K. An analysis of the proof of Theorem 20.1.5 shows that each ˇ 2 K can be written as ˛ ˇD r with ˛ 2 RK and r 2 Z. These rings of algebraic integers share many properties with the rational integers. While there may not be unique factorization into primes there is always prime factorization. Theorem 20.1.8. Let K be an algebraic number field and RK its ring of integers. Then each ˛ 2 RK is either 0, a unit or can be factored into a product of primes. We stress again that the prime factorization need not be unique. However from the existence of a prime factorization we can mimic Euclid’s original proof of the infinitude of primes (see [35]) to obtain: Corollary 20.1.9. There exist infinitely many primes in RK for any algebraic number ring RK . Just as any algebraic number field is finite dimensional over Q we will see that each RK is of finite degree over Q. That is if K has degree n over Q we show that there exists !1 ; : : : ; !n in RK such that each ˛ 2 RK is expressible as ˛ D m1 !1 C    C mn !n where m1 ; : : : ; mn 2 Z. Definition 20.1.10. An integral basis for RK is a set of integers !1 ; : : : ; ! t 2 RK such that each ˛ 2 RK can be expressed uniquely as ˛ D m1 !1 C    C m t ! t where m1 ; : : : ; m t 2 Z. The finite degree comes from the following result that shows there does exist an integral basis (see [35]). Theorem 20.1.11. Let RK be the ring of integers in the algebraic number field K of degree n over Q. Then there exists at least one integral basis for RK .

298

20.2

Chapter 20 Integral and Transcendental Extensions

Integral ring extensions

We now extend the concept of an algebraic integer to general ring extensions. We first need the idea of an R-algebra where R is a commutative ring with identity 1 ¤ 0. Definition 20.2.1. Let R be a commutative ring with an identity 1 ¤ 0. An Ralgebra or algebra over R is a unitary R-module A in which there is an additional multiplication such that (1) A is a ring with respect to the addition and this multiplication (2) .rx/y D x.ry/ D r.xy/ for all r 2 R and x; y 2 A. As examples of R-algebras first consider R D K where K is a field and let A D Mn .K/ the set of all .n  n/-matrices over K. Then Mn .K/ is a K-algebra. Further the set of polynomials KŒx is also a K-algebra. We now define ring extensions. Let A be a ring, not necessarily commutative, with an identity 1 ¤ 0, and R be a commutative subring of A which contains 1. Assume that R is contained in the center of A, that is, rx D xr for all r 2 R and x 2 A. We then call A a ring extension of R and write AjR. If AjR is a ring extension then A is an R-algebra in a natural manner. Let A be an R-algebra with an identity 1 ¤ 0. Then we have the canonical ring homomorphism  W R ! A, r 7! r  1. The image R0 WD .R/ is a subring of the center of A, and R 0 contains the identity element of A. Then AjR0 is a ring extension (in the above sense). Hence, if A is a R-algebra with an identity 1 ¤ 0 then we may consider R as a subring of A and AjR as a ring extension. We now will extend to the general context of ring extensions the ideas of integral elements and integral extensions. As above, let R be a commutative ring with an identity 1 ¤ 0 and let A be an R-algebra. Definition 20.2.2. An element a 2 A is said to be integral over R or integrally dependent over R if there is a monic polynomial f .x/ D x n C ˛n1 x n1 C    C ˛0 2 RŒx of degree n  1 over R with f .a/ D an C ˛n1 an1 C    C ˛0 D 0. That is, a is integral over R if it is a root of a monic polynomial of degree  1 over R. An equation that an integral element satisfies is called integral P equation of a over R. If A has an identity 1 ¤ 0 then we may write a0 D 1 and niD0 ˛i ai with ˛n D 1. Example 20.2.3. 1. Let EjK be a field extension. a 2 E is integral over K if and only if a is algebraic over K. If K is the quotient field of an integral domain R and a 2 E is algebraic over K then there exists an ˛ 2 R with ˛a integral over R, because if 0 D ˛n an C    C ˛0 then 0 D .˛n a/n C    C ˛nn1 ˛0 . 2. The elements of C which are integral over Z are precisely the algebraic integers over Z, that is, the roots of monic polynomials over Z.

299

Section 20.2 Integral ring extensions

Theorem 20.2.4. Let R be as above and A an R-algebra with an identity 1 ¤ 0. If A is, as an R-module, finitely generated then each element of A is integral over R. Proof. Let ¹b1 ; : : : ; bn º be a finite generating system of A, as an R-module. We may assume that b1 D 1, otherwise add 1 to the system. As explained in the preliminaries, without loss of generality, wePmay assume that R  A. Let a 2 A. For each 1  j  n we have an equation abj D nkD1 ˛kj bk for some ˛kj 2 R. In other words: n X

.˛kj  ıj k a/bk D 0

(??)

kD1

for j D 1; : : : ; n, where

´ ıj k D

0 if j ¤ k; 1 if j D k:

Define j k WD ˛kj  ıj k a and C D . j k /j;k . C is an .n  n/-matrix over the commutative ring RŒa; recall that RŒa has an identity element. Let CQ D . Qj k /j;k be the complimentary matrix of C . Then CQ C D .det C /En . From (??) we get 0D

n X j D1

Qij

X n



j k bk

kD1

D

n n X X kD1 j D1

Qij j k bk D

n X

.det C /ıik bk D .det C /bi

kD1

for all 1  i  n. Since b1 D 1 we have necessarily that det C D det.˛j k  ıj k a/j;k D 0 (recall that ıj k D ıkj ). Hence a is a root of the monic polynomial f .x/ D det.ıj k x  ˛j k / 2 RŒx of degree n  1. Hence a is integral over R. Definition 20.2.5. A ring extension AjR is called an integral extension if each element of A is integral over R. A ring extension AjR is called finite if A, as a R-module, is finitely generated. Recall that finite field extensions are algebraic extensions. As an immediate consequence of Theorem 20.2.4 we get the corresponding result for ring extensions. Theorem 20.2.6. Each finite ring extension AjR is an integral extension. Theorem 20.2.7. Let A be an R-algebra with an identity 1 ¤ 0. If a 2 A then the following are equivalent: (1) a is integral over R. (2) The subalgebra RŒa is, as an R-module, finitely generated. (3) There exists a subalgebra A0 of A which contains a and which is, as an Rmodule, finitely generated. A subalgebra of an algebra over R is a submodule which is also a subring.

300

Chapter 20 Integral and Transcendental Extensions

Proof. (1) ) (2): We have RŒa D ¹g.a/ W g 2 RŒxº. Let f .a/ D 0 be an integral equation of a over R. Since f is monic, by the division algorithm, for each g 2 RŒx there are h; r 2 RŒx with g D h  f C r and r D 0 or r ¤ 0 and deg.r/ < deg.f / DW n. Let r ¤ 0. Since g.a/ D r.a/, we get that ¹1; a; : : : ; an1 º is a generating system for the R-module RŒa. (2) ) (3): Take A0 D RŒa. (3) ) (1): Use Theorem 20.2.4 for A0 . For the remainder of this chapter all rings are commutative with an identity 1 ¤ 0. Theorem 20.2.8. Let AjR and BjA be finite ring extensions. Then also BjR is finite. Proof. From A D Re1 C    C Rem and B D Af1 C    C Afn we get B D Re1 f1 C    C Rem fn . Theorem 20.2.9. Let AjR be a ring extension. Then the following are equivalent: (1) There are finitely many, over R integral elements a1 ; : : : ; am in A such that A D RŒa1 ; : : : ; am . (2) AjR is finite. Proof. (2) ) (1): We only need to take for a1 ; : : : ; am a generating system of A as an R-module, and the result holds because A D Ra1 C    C Ram , and each ai is integral over R by Theorem 20.2.4. (1) ) (2): We use induction for m. If m D 0 then there is nothing to prove. Now let m  1, and assume that (1) holds. Define A0 D RŒa1 ; : : : ; am1 . Then A D A0 Œam , and am is integral over A0 . AjA0 is finite by Theorem 20.2.7. By the induction assumption, A0 jR is finite. Then AjR is finite by Theorem 20.2.8. Definition 20.2.10. Let AjR be a ring extension. Then the subset C D ¹a 2 A W a is integral over Rº  A is called the integral closure of R in A. Theorem 20.2.11. Let AjR be a ring extension. Then the integral closure of R in A is a subring of A with R  A. Proof. R  C because ˛ 2 R is a root of the polynomial x  ˛. Let a; b 2 C . We consider the subalgebra RŒa; b of the R-algebra A. RŒa; bjR is finite by Theorem 20.2.9. Hence, by Theorem 20.2.4, all elements from RŒa; b are integral over R, that is, RŒa; b  C . Especially, a C b, a  b and ab are in C . We extend to ring extensions the idea of a closure. Definition 20.2.12. Let AjR a ring extension. R is called integrally closed in A, if R itself is its integral closure in R, that is, R D C , the integral closure of R in A.

Section 20.2 Integral ring extensions

301

Theorem 20.2.13. For each ring extension AjR the integral closure C of R in A is integrally closed in A. Proof. Let a 2 A be integral over C . Then an C ˛n1 an1 C    C ˛0 D 0 for some ˛i 2 C , n  1. Then a is also integral over the R-subalgebra A0 D RŒ˛0 ; : : : ; ˛n1  of C ; and A0 jR is finite. Further A0 ŒajA is finite. Hence A0 ŒajR is finite. By Theorem 20.2.4, then a 2 A0 Œa is already integral over R, that is, a 2 C . Theorem 20.2.14. Let AjR and BjA be ring extensions. If AjR and BjA are integral extensions then also BjR is an integral extension (and certainly vice versa). Proof. Let C be the integral closure of R in B. We have A  C since AjR is integral. Together with BjA we also have that BjC is integral. By Theorem 20.2.13 we get that C is integrally closed in B. Hence, B D C . We now consider integrally closed integral domains. Definition 20.2.15. An integral domain R is called integrally closed if R is integrally closed in its quotient field K. Theorem 20.2.16. Each unique factorization domain R is integrally closed. Proof. Let ˛ 2 K and ˛ D ab with a; b 2 R, a ¤ 0. Since R is a unique factorization domain we may assume that a and b are relatively prime. Let ˛ be integral over R. Then we have over R an integral equation ˛ n C an1 ˛ n1 C    C a0 D 0 for ˛. Multiplication with b n gives an C ban1 C    C b n a0 D 0. Hence b is a divisor of an . Since a and b are relatively prime in R, we have that b is a unit in R and, hence, ˛ D ab 2 R. Theorem 20.2.17. Let R be an integral domain and K its quotient field. Let EjK be a finite field extension. Let R be integrally closed, and ˛ 2 E be integral over R. Then the minimal polynomial g 2 KŒx of ˛ over K has only coefficients of R. Proof. Let g 2 KŒx be the minimal polynomial of ˛ over K (recall that g is monic by definition). Let EN be an algebraic closure of E. Then g.x/ D .x  ˛1 /    .x  ˛n / N There are K-isomorphisms i W K.˛/ ! EN with i .˛/ D ˛i . with ˛1 D ˛ over E. Hence all ˛i are also integral over R. Since all coefficients of g are polynomial expressions Cj .˛1 ; : : : ; ˛n / in the ˛i we get that all coefficients of g are integral over R (see Theorem 20.2.11). Now g 2 RŒx because g 2 KŒx and R is integrally closed. Theorem 20.2.18. Let R be an integrally closed integral domain and K be its quotient field. Let f; g; h 2 KŒx be monic polynomials over K with f D gh. If f 2 RŒx then also g; h 2 RŒx.

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Chapter 20 Integral and Transcendental Extensions

Proof. Let E be the splitting field of f over K. Over E we have f .x/ D .x  ˛1 /    .x  ˛n /. Since f is monic all ˛k are integral over R (see the proof Q of Theorem 20.2.17).Q Since f D gh there are I; J  ¹1; : : : ; nº with g.x/ D i 2I .x  ˛i / and h.x/ D j 2J .x  ˛j /. As polynomial expressions in the ˛i , i 2 I , and ˛j , j 2 J , respectively, the coefficients of g and h, respectively, are integral over R. On the other side all these coefficients are in K and R is integrally closed. Hence g; h 2 RŒx. Theorem 20.2.19. Let EjR be an integral ring extension. If E is a field then also R is a field. Proof. Let ˛ 2 R n ¹0º. The element ˛1 2 E satisfies an integral equation . ˛1 /n C an1 . ˛1 /n1 C    C a0 D 0 over R. Multiplication with ˛n1 gives ˛1 D an1  an2 ˛      a0 ˛ n1 2 R. Hence, R is a field.

20.3

Transcendental field extensions

Recall that a transcendental number is an element of C that is not algebraic over Q. More generally if EjK is a field extension then an element ˛ 2 E is transcendental over K if it is not algebraic, that is, it is not a zero of any polynomial f .x/ 2 KŒx. Since finite extensions are algebraic clearly EjK will contain transcendental elements only if ŒE W K D 1. However this is not sufficient. The field A of algebraic numbers is algebraic over Q but infinite dimensional over Q. We now extend the idea of a transcendental number to that of a transcendental extension. Let K  E be fields, that is, EjK is a field extension. Let M be a subset of E. The algebraic cover of M in E is defined to be the algebraic closure H.M / of K.M / in E, that is, HK;E .M / D H.M / D ¹˛ 2 E W ˛ algebraic over K.M /º. H.M / is a field with K  K.M /  H.M /  E. ˛ 2 E is called algebraically dependent on M (over K) if ˛ 2 H.M /, that is, if ˛ is algebraic over K.M /. The following are clear. 1. M  H.M /, 2. M  M 0 ) H.M /  H.M 0 / and 3. H.H.M // D H.M /. Definition 20.3.1. (a) M is said to be algebraically independent (over K) if ˛ … H.M n ¹˛º/ for all ˛ 2 M , that is, if each ˛ 2 M is transcendental over K.M n ¹˛º/. (b) M is said to be algebraically dependent (over K) if M is not algebraically independent. The proofs of the statements in the following lemma are straightforward.

Section 20.3 Transcendental field extensions

303

Lemma 20.3.2. (1) M is algebraically dependent if and only if there exists an ˛ 2 M which is algebraic over K.M n ¹˛º/. (2) Let ˛ 2 M . Then ˛ 2 H.M n ¹˛º/ , H.M / D H.M n ¹˛º/. (3) If ˛ … M and ˛ is algebraic over K.M / then M [ ¹˛º is algebraically dependent. (4) M is algebraically dependent if and only if there is a finite subset in M which is algebraically dependent. (5) M is algebraically independent if and only if each finite subset of M is algebraically independent. (6) M is algebraically independent if and only if the following holds: If ˛1 ; : : : ; ˛n are finitely many, pairwise different elements of M then the canonical homomorphism  W KŒx1 ; : : : ; xn  ! E, f .x1 ; : : : ; xn / 7! f .˛1 ; : : : ; ˛n / is injective; or in other words: for all f 2 KŒx1 ; : : : ; xn  we have that f D 0 if f .˛1 ; : : : ; ˛n / D 0, that is, there is no nontrivial algebraic relation between the ˛1 ; : : : ; ˛n over K. (7) Let M  E, ˛ 2 E. If M is algebraically independent and M [ ¹˛º algebraically dependent then ˛ 2 H.M /, that is, ˛ is algebraically dependent on M . (8) Let M  E, B  M . If B is maximal algebraically independent, that is, if ˛ 2 M n B then B [ ¹˛º is algebraically dependent, then M  H.B/, that is, each element of M is algebraic over K.B/. We will show that any field extension can be decomposed into a transcendental extension over an algebraic extension. We need the idea of a transcendence basis. Definition 20.3.3. B  E is called a transcendence basis of the field extension EjK if the following two conditions are satisfied: 1. E D H.B/, that is, the extension EjK.B/ is algebraic. 2. B is algebraically independent over K. Theorem 20.3.4. If B  E then the following are equivalent: (1) B is a transcendence basis of EjK. (2) If B  M  E with H.M / D E, then B is a maximal algebraically independent subset of M . (3) There exists a subset M  E with H.M / D E which contains B as a maximal algebraically independent subset.

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Chapter 20 Integral and Transcendental Extensions

Proof. (1) ) (2): Let ˛ 2 M n B. We have to show that B [ ¹˛º is algebraically dependent. But this is clear because ˛ 2 H.B/ D E. (2) ) (3): We just take M D E. (3) ) (1): We have to show that H.B/ D E. Certainly M  H.B/. Hence, E D H.M /  H.H.B// D H.B/  E. We next show that any field extension does have a transcendence basis. Theorem 20.3.5. Each field extension EjK has a transcendence basis. More concretely: If there is a subset M  E such that EjK.M / is algebraic and if there is a subset C  M which is algebraically independent then there exists a transcendence basis B of EjK with C  B  M . Proof. We have to extend C to a maximal algebraically independent subset B of M . By Theorem 20.3.4, such a B is a transcendence basis of EjK. If M is finite then such a B certainly exists. Now let M be not finite. We argue analogously as for the existence of a basis of a vector space, for instance with Zorn’s lemma: If a partially ordered, nonempty set S is inductive, then there exist maximal elements in S. Here, a partially ordered, nonempty set S is said to be inductive if every totally ordered subset of S has an upper bound in S . The set N of all algebraically independent subsets of M which contain C is partially ordered with respect to “”, and N ¤ ; because C 2 N . Let K ¤ ; be an ascending chain in S N , that is, given an ascending chain ; ¤ Y1  Y2     . in N . The union U D Y 2K Y is also algebraically independent. Hence, there exists a maximal algebraically independent subset B  M with C  B. Theorem 20.3.6. Let EjK be a field extension and M be a subset of E for which EjK.M / is algebraic. Let C be an arbitrary subset of E which is algebraically independent on K. Then there exists a subset M 0  M with C \ M 0 D ; such that C [ M 0 is a transcendence basis of EjK. Proof. Take M [ C and define M 0 WD B n C in Theorem 20.3.5. Theorem 20.3.7. Let B; B 0 be two transcendence bases of the field extension EjK. Then there is a bijection  W B ! B 0 . In other words, any two transcendence bases of EjK have the same cardinal number. Proof. (a) If B is a transcendental basis of EjK S and M is a subset of E such that EjK.M / is algebraic then we may write B D ˛2M B˛ with finite sets B˛ . Especially, if B is infinite then the cardinal number of B is not bigger than the cardinal number of M . (b) Let B and B 0 be two transcendence bases of EjK. If B and B 0 are both infinite then B and B 0 have the same cardinal number by (a) and the theorem by

Section 20.3 Transcendental field extensions

305

Schroeder–Bernstein [5]. We now prove Theorem 20.3.7 for the case that EjK has a finite transcendence basis. Let B be finite with n elements. Let C be an arbitrary algebraically independent subset in E over K with m elements. We show that m  n. Let C D ¹˛1 ; : : : ; ˛m º with m  n. We show by induction that for each integer k, 0  k  n, there are subsets B ¥ B1 ¥    ¥ Bk of B such that ¹˛1 ; : : : ; ˛k º [ Bk is a transcendence basis of EjK and ¹˛1 ; : : : ; ˛k º \ Bk D ;. For k D 0 we take B0 D B, and the statement holds. Assume now that the statement is correct for 0  k < n. By Theorem 20.3.4 and 20.3.5 there is a subset BkC1 of ¹˛1 ; : : : ; ˛k º [ Bk such that ¹˛1 ; : : : ; ˛kC1 º [ BkC1 is a transcendence basis of EjK and ¹˛1 ; : : : ; ˛kC1 º \ BkC1 D ;. Then necessarily BkC1  Bk . Assume Bk D BkC1 . Then on one side, Bk [ ¹˛1 ; : : : ; ˛kC1 º is algebraic independent because Bk D BkC1 . On the other side, also Bk [¹˛1 ; : : : ; ˛k º[¹akC1 º is algebraically dependent, which gives a contradiction. Hence, BkC1 ¦ Bk . Now Bk has at most n  k elements, hence Bn D ;, that is, ¹˛1 ; : : : ; ˛n º D ¹˛1 ; : : : ; ˛n º [ Bn is a transcendence basis of EjK. Because C D ¹˛1 ; : : : ; ˛m º is algebraically independent, we cannot have m > n. Hence m  n; and B and B 0 have the same number of elements because B 0 must also be finite. Since the cardinality of any transcendence basis for a field extension EjK is the same we can define the transcendence degree. Definition 20.3.8. The transcendence degree trgd.EjK/ of a field extension is the cardinal number of one (and hence of each) transcendence basis of EjK. A field extension EjK is called purely transcendental, if EjK has a transcendence basis B with E D K.B/. We note the following facts: (1) If EjK is purely transcendental and B D ¹˛1 ; : : : ; ˛n º is a transcendence basis of EjK then E is K-isomorphic to the quotient field of the polynomial ring KŒx1 ; : : : ; xn  of the independence indeterminates x1 ; : : : ; xn . (2) K is algebraically closed in E if EjK is purely transcendental. (3) By Theorem 20.3.4, the field extension EjK has an intermediate field F , K  F  E, such that F jK is purely transcendental and EjF is algebraic. Certainly F is not uniquely determined. For example take Q  F  Q.i; /, and for F we may take F D Q./ and also F D Q.i/, for instance. (4) trgd.RjQ/ D trgd.CjQ/ D card R, the cardinal number of R. This holds because the set of the algebraic numbers (over Q) is countable. Theorem 20.3.9. Let EjK a field extension and F an arbitrary intermediate field, K  F  E. Let B a transcendence basis of F jK and B 0 a transcendence base of

306

Chapter 20 Integral and Transcendental Extensions

EjF . Then B \ B 0 D ;, and B [ B 0 is a transcendence basis of EjK. Especially: trgd.EjK/ D trgd.EjF / C trgd.F jK/. Proof. (1) Assume ˛ 2 B \ B 0 . As an element of F , then ˛ is algebraic over F .B 0 / n ¹˛º. But this gives a contradiction because ˛ 2 B 0 , and B 0 is algebraically independent over F . (2) F jK.B/ is an algebraic extension, and also F .B 0 /jK.B [ B 0 / D K.B/.B 0 /. Since the relation “algebraic extension” is transitive, we have that EjK.B [ B 0 / is algebraic. (3) Finally we have to show that B [ B 0 is algebraically independent over K. By Theorems 20.3.5 and 20.3.6 there is a subset B 00 of B [ B 0 with B \ B 00 D ; such that B [ B 00 is a transcendence basis of EjK. We have B 00  B 0 , and have to show that B 0  B 00 . Assume that there is an ˛ 2 B 0 with ˛ … B 00 . Then ˛ is algebraic over K.B [ B 00 / D K.B/.B 00 / and, hence, algebraic over F .B 00 /. Since B 00  B 0 we have that ˛ is algebraically independent over F , which gives a contradiction. Hence B 00 D B 0 . Theorem 20.3.10 (Noether’s normalization theorem). Let K be a field and A D KŒa1 ; : : : ; an . Then there exist elements u1 ; : : : ; um , 0  m  n, in A with the following properties: (1) KŒu1 ; : : : ; um  is K-isomorphic to the polynomial ring KŒx1 ; : : : ; xm  of the independent indeterminates x1 ; : : : ; xm . (2) The ring extension AjKŒu1 ; : : : ; um  is an integral extension, that is, for each a 2 AnKŒu1 ; : : : ; um  there exists a monic polynomial f .x/ D x n C˛n1 x n1 C    C ˛0 2 KŒu1 ; : : : ; um Œx of degree n  1 with f .a/ D an C ˛n1 an1 C    C ˛0 D 0. Especially AjKŒu1 ; : : : ; um  is finite. Proof. Without loss of generality, let the a1 ; : : : ; an be pairwise different. We prove the theorem by induction on n. If n D 1 then there is nothing to show. Now, let n  2, and assume that the statement holds for n1. If there is no nontrivial algebraic relation f .a1 ; : : : ; an / D 0 over K between the a1 ; : : : ; an then there is nothing to show. Hence, let there exists aP polynomial f 2 KŒx1 ; : : : ; xn  with f ¤ 0 and 1 n f .a1 ; : : : ; an / D 0. Let f D D.1 ;:::;n / c x1    xn . Let 2 ; 3 ; : : : ; n be   natural numbers which we specify later. Define b2 D a2  a1 2 , b3 D a3  a1 3 ,   . . . ; bn D an  a1 n . Then ai D bi C a1 i for 2  i  n, hence, f .a1 ; b2 C 2 n a1 ; : : : ; bn C a1 / D 0. We write R WD KŒx1 ; : : : ; xn  and consider the polynomial ring RŒy2 ; : : : ; yn  of the n  1 independent indeterminates y2 ; : : : ; yn over R. In   RŒy2 ; : : : ; yn  we consider the polynomial f .x1 ; y2 C x1 2 ; : : : ; yn C x1 n /. We may rewrite this polynomial as X  C  CCn n c x11 2 2 C g.x1 ; y2 ; : : : ; yn / D.1 ;:::;n /

Section 20.4 The transcendence of e and 

307

with a polynomial g.x1 ; y2 ; : : : ; yn / for which, asPa polynomial in x1 over KŒy2 ; : : : ;  C  CCn n , yn , the degree in x1 is smaller than the degree of D.1 ;:::;n / c x11 2 2 provided that we may choose the 2 ; : : : ; n in such a way that this really holds. We now specify the 2 ; : : : ; n . We write  WD .1; 2 ; : : : ; n / and define the scalar product  D 1  1 C 2 2 C    C n n . Choose p 2 N with p > deg.f / D max¹1 C    C n W c ¤ 0º. We now take  D .1; p; p 2 ; : : : ; p n1 /. If  D .1 ; : : : ; n / with c ¤ 0 and  0 D .10 ; : : : ; n0 / with c0 ¤ 0 are different n-tuple then indeed  ¤  0 because i ; i0 < p for all i, 1  i  n. This follows from the uniqueness of the p-adic expression of a natural number. Hence, we may choose   2 ; : : : ; n such that f .x1 ; y2 C x1 2 ; : : : ; yn C x1 n / D cx1N C h.x1 ; y2 ; : : : ; yn / with c 2 K, c ¤ 0, and h 2 KŒy2 ; : : : ; yn Œx1  has in x1 a degree < N . If we divide by c and take a1 ; b2 ; : : : ; bn for x1 ; y2 ; : : : ; yn then we get an integral equation of a1 over KŒb2 ; : : : ; bn . Therefore, the ring extension A D KŒa1 ; : : : ; an jKŒb2 ; : : : ; bn   is integral (see Theorem 20.2.9), ai D bi C a1 i for 2  i  n. By induction there exists elements u1 ; : : : ; um in KŒb2 ; : : : ; bn  with the following properties: 1. KŒu1 ; : : : ; um is a polynomial ring of the m independent indeterminates u1 ; : : : ; um and 2. KŒb2 ; : : : ; bn jKŒu1 ; : : : ; um  is integral. Hence, also AjKŒu1 ; : : : ; um  is integral by Theorem 20.2.14. Corollary 20.3.11. Let EjK be a field extension. If E D KŒa1 ; : : : ; an  for a1 ; : : : ; an 2 E then EjK is algebraic. Proof. By Theorem 20.3.10 we have that E contains a polynomial ring KŒu1 ; : : : ; um , 0  m  n, of the m independent indeterminates u1 ; : : : ; um as a subring for which EjKŒu1 ; : : : ; um  is integral. We claim that then already KŒu1 ; : : : ; um  is a field. To prove that, let a 2 KŒu1 ; : : : ; um , a ¤ 0. The element a1 2 E satisfies an integral equation .a1 /n C ˛n1 .a1 /n1 C    C ˛0 D 0 over KŒu1 ; : : : ; um  DW R. Hence, a1 D ˛n1  ˛n2 a      ˛0 an1 2 R. Therefore R is a field which proves the claim. This is possible only for m D 0, and then EjK is integral, that is here algebraic.

20.4

The transcendence of e and 

Although we have shown that within C there are continuously many transcendental numbers we have only shown that one particular number is transcendental. In this section we prove that the numbers e and  are transcendental. We start with e. Theorem 20.4.1. e is a transcendental number, that is, transcendental over Q.

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Chapter 20 Integral and Transcendental Extensions

Proof. Let f .x/ 2 RŒx with the degree of f .x/ D m  1. Let z1 2 C, z1 ¤ 0, and

W Œ0; 1 ! C, .t / D t z1 . Let  Z z1  Z z1 z I.z1 / D e f .z/dz D e z1 z f .z/dz: 

0



R z1

By . 0 / we mean the integral from 0 to z1 along . Recall that  Z z1   Z z1  e z1 z f .z/dz D f .z1 / C e z1 f .0/ C e z1 z f 0 .z/dz: 0

0





It follows then by repeated partial integration that P P (1) I.z1 / D e z1 jmD0 f .j / .0/  jmD0 f .j / .z1 /. Let jf j.x/ be the polynomial that we get if we replace the coefficients of f .x/ by their absolute values. Since je z1 z j  e jz1 zj  e jz1 j , we get (2) jI.z1 /j  jz1 je jz1 j jf j.jz1 j/. Now assume that e is an algebraic number, that is, (3) q0 C q1 e C    C qn e n D 0 for n  1 and integers q0 ¤ 0; q1 ; : : : ; qn , and the greatest common divisor of q0 ; q1 ; : : : ; qn , is equal to 1. We consider now the polynomial f .x/ D x p1 .x  1/p : : : .x  n/p with p a sufficiently large prime number, and we consider I.z1 / with respect to this polynomial. Let J D q0 I.0/ C q1 I.1/ C    C qn I.n/: From (1) and (3) we get that J D

m X n X

qk f .j / .k/;

j D0 kD0

P where m D .n C 1/p  1 since .q0 C q1 e C    C qn e n /. jmD0 .f .j / .0// D 0. Now, f .j / .k/ D 0 if j < p, k > 0, and if j < p  1 then k D 0, and hence .j / .k/ is an integer that is divisible by pŠ for all j; k except for j D p  1, k D 0. f Further, f .p1/ .0/ D .p  1/Š.1/np .nŠ/p , and hence, if p > n, then f .p1/ .0/ is an integer divisible by .p  1/Š but not by pŠ. It follows that J is a nonzero integer that is divisible by .p  1/Š if p > jq0 j and p > n. So let p > n; p > jq0 j, so that jJ j  .p  1/Š. Now, jf j.k/  .2n/m . Together with (2) we then get that jJ j  jq1 jejf j.1/ C    C jqn jne n jf j.n/  c p for a number c independent of p. It follows that .p  1/Š  jJ j  c p ;

309

Section 20.4 The transcendence of e and 

that is, 1

jJ j c p1 c : .p  1/Š .p  1/Š

This gives a contradiction, since dental.

c p1 .p1/Š

! 0 as p ! 1. Therefore, e is transcen-

We now move on to the transcendence of . We first need the following lemma. Lemma 20.4.2. Suppose ˛ 2 C is an algebraic number and f .x/ D an x n C  Ca0 , n  1, an ¤ 0, and all ai 2 Z .f .x/ 2 ZŒx/ with f .˛/ D 0. Then an ˛ is an algebraic integer. Proof. ann1 f .x/ D ann x n C ann1 an1 x n1 C    C ann1 a0 D .an x/n C an1 .an x/n1 C    C ann1 a0 D g.an x/ D g.y/ 2 ZŒy where y D an x and g.y/ is monic. Then g.an ˛/ D 0, and hence an ˛ is an algebraic integer. Theorem 20.4.3.  is a transcendental number, that is, transcendental over Q. Proof. Assume that  is an algebraic number. Then  D i  is also algebraic. Let 1 D ; 2 ; : : : ; d be the conjugates of . Suppose p.x/ D q0 C q1 x C    C qd x d 2 ZŒx;

qd > 0;

and

gcd.q0 ; : : : ; qd / D 1

is the entire minimal polynomial of  over Q. Then 1 D ; 2 ; : : : ; d are the zeros of this polynomial. Let t D qd . Then from Lemma 20.4.2, t i is an algebraic integer for all i. From e i C 1 D 0 and from 1 D i  we get that .1 C e 1 /.1 C e 2 /    .1 C e d / D 0: The product on the left side can be written as a sum of 2d terms e  , where  D 1 1 C    C d d , j D 0 or 1. Let n be the number of terms 1 1 C    C d d that are nonzero. Call these ˛1 ; : : : ; ˛n . We then have an equation q C e ˛1 C    C e ˛n D 0 with q D 2d  n > 0. Recall that all t ˛i , are algebraic integers and we consider the polynomial f .x/ D t np x p1 .x  ˛1 /p    .x  ˛n /p

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Chapter 20 Integral and Transcendental Extensions

with p a sufficiently large prime integer. We have f .x/ 2 RŒx, since the ˛i are algebraic numbers and the elementary symmetric polynomials in ˛1 ; : : : ; ˛n are rational numbers. Let I.z1 / be defined as in the proof of Theorem 20.4.1, and now let J D I.˛1 / C    C I.˛n /: From (1) in the proof of Theorem 20.4.1 and (4) we get J D q

m X j D0

f .j / .0/ 

n m X X

f .j / .˛k /;

j D0 kD1

with m D P.n C 1/p  1. Now, nkD1 f .j / .˛k / is a symmetric polynomial in t ˛1 ; : : : ; t ˛n with integer coefalgebraic ficients since the t ˛i are P P integers. It follows from the main theorem on symmetric polynomials that jmD0 nkD1 f .j / .˛k / is an integer. Further, f .j / .˛k / D 0 P P for j < p. Hence jmD0 nkD1 f .j / .˛k / is an integer divisible by pŠ. Now, f .j / .0/ is an integer divisible by pŠ if j ¤ p  1, and f .p1/ .0/ D .p  1/Š.t /np .˛1 : : : ˛n /p is an integer divisible by .p  1/Š but not divisible by pŠ if p is sufficiently large. In particular, this is true if p > jt n .˛1    ˛n /j and also p > q. From (2) in the proof of Theorem 20.4.1 we get that jJ j  j˛1 je j˛1 j jf j.j˛1 j/ C    C j˛n je j˛n j jf j.j˛n j/  c p for some number c independent of p. As in the proof of Theorem 20.4.1, this gives us .p  1/Š  jJ j  c p ; that is, 1

c p1 jJ j c : .p  1/Š .p  1/Š

This, as before, gives a contradiction, since is transcendental.

20.5

c p1 .p1/Š

! 0 as p ! 1. Therefore, 

Exercises

1. A polynomial p.x/ 2 ZŒx is primitive if the GCD of all its coefficients is 1. Prove the following: (i) If f .x/ and g.x/ are primitive then so is f .x/g.x/. (ii) If f .x/ 2 ZŒx is monic then it is primitive.

311

Section 20.5 Exercises

(iii) If f .x/ 2 QŒx then there exists a rational number c such that f .x/ D cf1 .x/ with f1 .x/ primitive. p 2. Let d be a square-free integer and K D Q. d / be a quadratic field. Let RK be the subring of K of the algebraic integers of K. Show that p (i) RKpD ¹m C n d W m; n 2 Zº if d  2.mod 4/ or d  3.mod 4/. ¹1; d º is an integral basis for RK . p

(ii) RK D ¹m C n 1C2 basis for RK .

d

p

W m; n 2 Zº if d  1. mod 4/. ¹1; 1C2

d

º is an integral

(iii) If d < 0 then there are only finitely many units in RK . (iv) If d > 0 then there are infinitely many units in RK . 3. Let K D Q.˛/ with ˛ 3 C ˛ C 1 D 0 and RK the subring of the algebraic integers in K. Show that (i) ¹1; ˛; ˛ 2 º is an integral basis for RK . (ii) RK D ZŒ˛. 4. Let AjR be an integral ring extension. If A is an integral domain and R a field then A is also a field. 5. Let AjR be an integral extension. Let P be a prime ideal of A and p be a prime ideal of R such that P \ R D p. Show that (i) If p is maximal in R then P is maximal in A. (Hint: consider A=P .) (ii) If P0 is another prime ideal of A with P0 \ R D p and P0  P then P D P0 . (Hint: we may assume that A is an integral domain and P \ R D ¹0º, otherwise go to A=P .) 6. Show that for a field extension EjK the following are equivalent: (i) ŒE W K.B/ < 1 for each transcendence basis B of EjK. (ii) trgd.EjK/ < 1 and ŒE W K.B/ < 1 for each transcendence basis B of EjK. (iii) There is a finite transcendence basis B of EjK with ŒE W K.B/ < 1. (iv) There are finitely many x1 ; : : : ; xn 2 E with E D K.x1 ; : : : ; xn /. 7. Let EjK be a field extension. If EjK is purely transcendental then K is algebraically closed in E.

Chapter 21

The Hilbert Basis Theorem and the Nullstellensatz

21.1

Algebraic Geometry

An extremely important application of abstract algebra and an application central to all of mathematics is the subject of algebraic geometry. As the name suggests this is the branch of mathematics that uses the techniques of abstract algebra to study geometric problems. Classically algebraic geometry involved the study of algebraic curves which roughly are the sets of zeros of a polynomial or set of polynomials in several variables over a field. For example, in two variables a real algebraic plane curve is the set of zeros in R2 of a polynomial p.x; y/ 2 RŒx; y. The common planar curves such as parabolas and the other conic sections are all plane algebraic curves. In actual practice plane algebraic curves are usually considered over the complex numbers and are projectivized. The algebraic theory that deals most directly with algebraic geometry is called commutative algebra. This is the study of commutative rings, ideals in commutative rings and modules over commutative rings. A large portion of this book has dealt with commutative algebra. Although we will not consider the geometric aspects of algebraic geometry in general we will close the book by introducing some of the basic algebraic ideas that are crucial to the subject. These include the concept of an algebraic variety or algebraic set and its radical. We also state and prove two of the cornerstones of the theory as applied to commutative algebra – the Hilbert basis theorem and the Nullstellensatz. In this chapter we consider a fixed field extension C jK and the polynomial ring KŒx1 ; : : : ; xn  of the n independent indeterminates x1 ; : : : ; xn . Again, in this chapter we often use letters a; b; m; p; P; A; Q; : : : for ideals in rings.

21.2

Algebraic Varieties and Radicals

We first define the concept of an algebraic variety. Definition 21.2.1. If M  KŒx1 ; : : : ; xn  then we define N .M / D ¹.˛1 ; : : : ; ˛n / 2 C n W f .˛1 ; : : : ; ˛n / D 0 8f 2 M º: ˛ D .˛1 ; : : : ; ˛n / 2 N .M / is called a zero (Nullstelle) of M in C n ; and N .M / is called the zero set of M in C n . If we want to mention C then we write N .M / D

Section 21.2 Algebraic Varieties and Radicals

313

NC .M /. A subset V  C n of the form V D N .M / for some M  KŒx1 ; : : : ; xn  is called an algebraic variety or (affine) algebraic set of C n over K, or just an algebraic K-set of C n . For any subset N of C n we can reverse the procedure and consider the set of polynomials whose zero set is N . Definition 21.2.2. Suppose that N  C n . Then I.N / D ¹f 2 KŒx1 ; : : : ; xn  W f .˛1 ; : : : ; ˛n / D 0 8.˛1 ; : : : ; ˛n / 2 N º: Instead of f 2 I.N / we also say that f vanishes on N (over K). If we want to mention K then we write I.N / D IK .N /. What is important is that the set I.N / forms an ideal. The proof is straightforward. Theorem 21.2.3. For any subset N  C n the set I.N / is an ideal in KŒx1 ; : : : ; xn ; it is called the vanishing ideal of N  C n in KŒx1 ; : : : ; xn , The following result examines the relationship between subsets in C n and their vanishing ideals. Theorem 21.2.4. The following properties hold: (1) M  M 0 ) N .M 0 /  N .M /, (2) If a D .M / is the ideal in KŒx1 ; : : : ; xn  generated by M , then N .M / D N .a/, (3) N  N 0 ) I.N 0 /  I.N /, (4) M  I N .M / for all M  KŒx1 ; : : : ; xn , (5) N  N I.N / for all N  C n ,

T P (6) If .ai /P . i 2I ai /. i2I is a family of ideals in KŒx1 ; : : : ; xn  then i 2I N .ai /DNS Here i2I ai is the ideal in KŒx1 ; : : : ; xn , generated by the union i 2I ai , (7) If a; b are ideals in KŒx1 ; : : : ; xn  then N .a/ [ N .b/ D N .ab/ D N .a \ b/. Here ab is the ideal in KŒx1 ; : : : ; xn  generated by all products fg where f 2 a and g 2 b, (8) N .M / D N I N .M / for all M  KŒx1 ; : : : ; xn , (9) V D N I.V / for all algebraic K-sets V , (10) I.N / D I N I.N / for all N  C n . Proof. The proofs are straightforward. Hence, we prove only (7), (8) and (9). The rest can be left as exercise for the reader. Proof of (7): Since ab  a\b  a; b we have by (1) the inclusion N .a/[N .b/  N .a \ b/  N .ab/. Hence, we have to show that N .ab/  N .a/ [ N .b/.

314

Chapter 21 The Hilbert Basis Theorem and the Nullstellensatz

Let ˛ D .˛1 ; : : : ; ˛n / 2 C n be a zero of ab but not a zero of a. Then there is an f 2 a with f .˛/ ¤ 0, and hence for all g 2 b we get f .˛/g.˛/ D .fg/.˛/ D 0 and, hence, g.˛/ D 0. Therefore ˛ 2 N .b/. Proof of (8) and (9): Let M  KŒx1 ; : : : ; xn . Then, on the one side, M  I N .M / by (5) and further N I N .M /  N .M / by (1). On the other side, N .M /  N I N .M / by (6). Therefore N .M / D N I N .M / for all M  KŒx1 ; : : : ; xn . Now, the algebraic K-sets of C n are precisely the sets of the form V D N .M /. Hence, V D N I.V /. We make the following agreement: if a is an ideal in KŒx1 ; : : : ; xn  then we write a G KŒx1 ; : : : ; xn : If a G KŒx1 ; : : : ; xn  then we do not have a D I N .a/ in general, that is, a is in general not equal to the vanishing ideal of its zero set in C n . The reason for this is that not each ideal a occurs as a vanishing ideal of some N  C n . If a D I.N / then we must have: (?) f m 2 a; m  1 H) f 2 a: Hence, for instance, if a D .x12 ; : : : ; xn2 / G KŒx1 ; : : : ; xn  then a is not of the form a D I.N / for some N  C n . We now define the radical of an ideal. Definition 21.2.5. Let R be a commutative ring, and a G R be an ideal in R. Then p p a D ¹f 2 R W f m 2 a for some m 2 Nº is an ideal in R. a is called the radical p of a (in R). a is said to be reduced if a D a. p We note that the 0 is called the nil radical of R; it contains exactly the nilpotent elements of R, that is, the elements a 2 R with am D 0 for some m 2 N. p Let a G R be an ideal in R and  W R ! R=a the canonical mapping. Then a is exactly the preimage of the nil radical of R=a.

21.3

The Hilbert Basis Theorem

In this section we show that if K is a field then each ideal a G KŒx1 ; : : : ; xn  is finitely generated. This is the content of the Hilbert basis theorem. This has as an important consequence that any algebraic variety of C n is the zero set of only finitely many polynomials. The Hilbert basis theorem follows directly from the following Theorem 21.3.2. Before we state this theorem we need a definition. Definition 21.3.1. Let R be a commutative ring with an identity 1 ¤ 0. R is said to be noetherian if each ideal in R is generated by finitely many elements, that is, each ideal in R is finitely generated.

Section 21.4 The Hilbert Nullstellensatz

315

Theorem 21.3.2. Let R be a noetherian ring. Then the polynomial ring RŒx over R is also noetherian. Proof. Let 0 ¤ fk 2 RŒx. With deg.fk / we denote the degree of fk . Let a G RŒx be an ideal in RŒx. Assume that a is not finitely generated. Then, especially, a ¤ 0. We construct a sequence of polynomials fk 2 a such that the highest coefficients ak generate an ideal in R which is not finitely generated. This produces then a contradiction, and, hence, a is in fact finitely generated. Choose f1 2 a, f1 ¤ 0, so that deg.f1 / D n1 is minimal. If k  1 then choose fkC1 2 a, fkC1 … .f1 ; : : : ; fk / so that deg.fkC1 / D nkC1 is minimal for the polynomials in a n .f1 ; : : : ; fk /. This is possible because we assume that a is not finitely generated. We have nk  nkC1 by our construction. Further .a1 ; : : : ; ak / ¦ .a1 ; : : : ; ak ; akC1 /. Proof of this claim: Assume that .a1 ; : : : ; ak / D .a1 ; : : : ; ak ; akC1 /. Then akC1 2 Pk .a1 ; : : : ; ak /. Hence, there are bi 2 R with akC1 D i D1 ai bi . Let g.x/ D Pk n n kC1 i , hence, g 2 .f1 ; : : : ; fk / and g D akC1 x nkC1 C    . ThereiD1 bi fi .x/x fore deg.fkC1  g/ < nkC1 and fkC1  g … .f1 ; : : : ; fk / which contradicts the choice of fkC1 . This proves the claim. Hence .a1 ; : : : ; ak / ¦ .a1 ; : : : ; ak ; akC1 / which contradicts the fact that R is noetherian. Hence a is finitely generated. We now have the Hilbert basis theorem. Theorem 21.3.3 (Hilbert basis theorem). Let K be a field. Then each ideal a G KŒx1 ; : : : ; xn  is finitely generated, that is, a D .f1 ; : : : ; fm / for finitely many f1 ; : : : ; fm 2 KŒx1 ; : : : ; xn . Corollary 21.3.4. If C jK is a field extension then each algebraic K-set V of C n is already the zero set of only finitely many polynomials f1 ; : : : ; fm 2 KŒx1 ; : : : ; xn : V D ¹.˛1 ; : : : ; ˛n / 2 C n W fi .˛1 ; : : : ; ˛n / D 0 for i D 1; : : : ; mº: Further we write V D N .f1 ; : : : ; fm /.

21.4

The Hilbert Nullstellensatz

Vanishing ideals of subsets of C n are not necessarily reduced. For an arbitrary field C , the condition f m 2 a; m  1 H) f 2 a is, in general, not sufficient for a G KŒx1 ; : : : ; xn  to be a vanishing ideal of a subset of C n . For example let n  2, K D C D R and a D .x12 C    C xn2 / G RŒx1 ; : : : ; xn . a is a prime ideal in RŒx1 ; : : : ; xn  because x12 C    C xn2 is a prime element in

316

Chapter 21 The Hilbert Basis Theorem and the Nullstellensatz

RŒx1 ; : : : ; xn . Hence, a is reduced. But, on the other side, N .a/ D ¹0º and I.¹0º/ D .x1 ; : : : ; xn /. Therefore a is not of the form I.N / for some N  C n . If this would be the case, then a D I.N / D I N I.N / D I ¹0º D .x1 ; : : : ; xn / because of Theorem 21.2.4(10), which gives a contradiction. The Nullstellensatz of Hilbert which we give in two forms shows that if a is rep duced, that is, a D a, then I N .a/ D a. Theorem 21.4.1 (Hilbert’s Nullstellensatz, first form). Let C jK be a field extension p with C algebraically closed. If a G KŒx1 ; : : : ; xn  then I N .a/ D a. Moreover, if a p is reduced, that is, a D a, then I N .a/ D a. Therefore N defines a bijective map between the set of reduced ideals in KŒx1 ; : : : ; xn  and the set of the algebraic K-sets in C n ; and I defines the inverse map. The proof follows from: Theorem 21.4.2 (Hilbert’s Nullstellensatz, second form). Let C jK be a field extension with C algebraically closed. Let a G KŒx1 ; : : : ; xn  with a ¤ KŒx1 ; : : : ; xn . Then there exists an ˛ D .˛1 ; : : : ; ˛n / 2 C n with f .˛/ D 0 for all f 2 a, that is, NC .a/ ¤ ;. Proof. Since a ¤ KŒx1 ; : : : ; xn  there exists a maximal ideal m G KŒx1 ; : : : ; xn  with a  m. We consider the canonical map  W KŒx1 ; : : : ; xn  ! KŒx1 ; : : : ; xn =m. Let ˇi D .xi / for i D 1; : : : ; n. Then KŒx1 ; : : : ; xn =m D KŒˇ1 ; : : : ; ˇn  DW E. Since m is maximal, E is a field. Moreover EjK is algebraic by Corollary 20.3.11. Hence there exists a K-homomorphism W KŒˇ1 ; : : : ; ˇn  ! C (C is algebraically closed). Let ˛i D .ˇi /; we have f .˛1 ; : : : ; ˛n / D 0 for all f 2 m. Since a  m this holds also for all f 2 a. Hence we get a zero .˛1 ; : : : ; ˛n / of a in C n . Proof of Theorem 21.4.1. Let a G KŒx1 ; : : : ; xn , and let f 2 I N .a/. We have to show that f m 2 a for some m 2 N. If f D 0 then there is nothing to show. Now, let f ¤ 0. We consider KŒx1 ; : : : ; xn  as a subring of KŒx1 ; : : : ; xn ; xnC1  of the n C 1 independent indeterminates x1 ; : : : ; xn ; xnC1 . In KŒx1 ; : : : ; xn ; xnC1  we consider the ideal aN D .a; 1  xnC1 f / G KŒx1 ; : : : ; xn ; xnC1 , generated by a and 1  xnC1 f . Case 1: aN ¤ KŒx1 ; : : : ; xn ; xnC1 . aN then has a zero .ˇ1 ; : : : ; ˇn ; ˇnC1 / in C nC1 by Theorem 21.2.4. Hence, for N we have the equations: .ˇ1 ; : : : ; ˇn ; ˇnC1 / 2 N .a/ (1) g.ˇ1 ; : : : ; ˇn / D 0 for all g 2 a and (2) f .ˇ1 ; : : : ; ˇn /ˇnC1 D 1. From (1) we get .ˇ1 ; : : : ; ˇn / 2 N .a/. Hence, especially, f .ˇ1 ; : : : ; ˇn / D 0 for our f 2 I N .a/. But this contradicts (2). Therefore aN ¤ KŒx1 ; : : : ; xn ; xnC1  is not possible. Therefore we have

Section 21.5 Applications and Consequences of Hilbert’s Theorems

317

Case 2: aN D KŒx1 ; : : : ; xn ; xnC1 , that is, 1 2 a. N Then there exists a relation of the form X hi gi C h.1  xnC1 f / for some gi 2 a and hi ; h 2 KŒx1 ; : : : ; xn ; xnC1 : 1D i

The map xi 7! xi for 1  i  n and xnC1 7! f1 defines a homomorphism  W KŒx1 ; : : : ; xn ; xnC1  !PK.x1 ; : : : ; xn /, the quotient field of KŒx1 ; : : : ; xn . From (3) we get a relation 1 D i hi .x1 ; : : : ; xn ; f1 /gi .x1 ; : : : ; xn / in K.x1 ; : : : ; xn /. If we P multiply this with a suitable power f m of f we get f m D i hQ i .x1 ; : : : ; xn /gi .x1 ; : : : ; xn / for some polynomials hQ 2 KŒx1 ; : : : ; xn . Since gi 2 a we get f m 2 a.

21.5

Applications and Consequences of Hilbert’s Theorems

Theorem 21.5.1. Each nonempty set of algebraic K-sets in C n contains a minimal element. In other words: For each descending chain V1 V2    Vm VmC1   

(1)

of algebraic K-sets Vi in C n there exists an integer m such that Vm D VmC1 D VmC2 D    , or equivalently, every strictly descending chain V1 ¥ V2 ¥    of algebraic K-sets Vi in C n is finite. Proof. We apply the operator I , that is, we pass to the vanishing ideals. This gives an ascending chain of ideals I.V1 /  I.V2 /      I.Vm /  I.VmC1 /     :

(2)

The union of the I.Vi / is an ideal in KŒx1 ; : : : ; xn , and, hence, by Theorem 21.3.3 finitely generated. Hence, there is an m with I.Vm / D I.VmC1 / D I.VmC2 / D    . Now we apply the operator N and get the desired result because Vi D N I.Vi / by Theorem 21.2.4 (10). Definition 21.5.2. An algebraic K-set V ¤ ; in C n is called irreducible if it is not describable as a union V D V1 [ V2 of two algebraic K-sets Vi ¤ ; in C n with Vi ¤ V for i D 1; 2. An irreducible algebraic K-set in C n is also called a K-variety in C n . Theorem 21.5.3. An algebraic K-set V ¤ ; in C n is irreducible if and only if its vanishing ideal Ik .V / D I.V / is a prime ideal of R D KŒx1 ; : : : ; xn  with I.V / ¤ R. Proof. (1) Let V be irreducible. Let fg 2 I.V /. Then V D N I.V /  N .fg/ D N .f / [ N .g/, hence V D V1 [ V2 with the algebraic K-sets V1 D N .f / \ V and

318

Chapter 21 The Hilbert Basis Theorem and the Nullstellensatz

V2 D N .g/ \ V . Now V is irreducible, hence, V D V1 or V D V2 ; say V D V1 . Then V  N .f /, and therefore f 2 I N .f /  I.V /. Since V ¤ ; we have further 1 … I.V /, that is, I.V / ¤ R. (2) Let I.V / G R with I.V / ¤ R be a prime ideal. Let V D V1 [ V2 , V1 ¤ V , with algebraic K-sets Vi in C n . First, I.V / D I.V1 [ V2 / D I.V1 / \ I.V2 / I.V1 /I.V2 /;

(?)

where I.V1 /I.V2 / is the ideal generated by all products fg with f 2 I.V1 /, g 2 I.V2 /. We have I.V1 / ¤ I.V / because otherwise V1 D N I.V1 / D N I.V / D V contradicting V1 ¤ V . Hence, there is a f 2 I.V1 / with f … I.V /. Now, I.V / ¤ R is a prime ideal, and hence, necessarily I.V2 /  I.V / by (?). It follows that V  V2 , and, hence, V is irreducible. Note that the affine space K n is, as the zero set of the zero polynomial 0, itself an algebraic K-set in K n . If K is infinite then I.K n / D ¹0º and, hence K n is irreducible by Theorem 21.5.3. Moreover, if K is infinite then K n can not be written as a union of finitely many proper algebraic K-subsets. If K is finite then K n is not irreducible. Further each algebraic K-set V in C n is also an algebraic C -set in C n . If V is an irreducible algebraic K-set in C n then, in general, it is not an irreducible algebraic C -set in C n . Theorem 21.5.4. Each algebraic K-set V in C n can be written as a finite union V D V1 [ V2 [    [ Vr of irreducible algebraic K-sets Vi in C n . If here Vi ª Vk for all pairs .i; k/ with i ¤ k then this presentation is unique, up to the ordering of the Vi ; and then the Vi are called the irreducible K-components of V . Proof. Let a be the set of all algebraic K-sets in C n which can not be presented as a finite union of irreducible algebraic K-sets in C n . Assume that a ¤ ;. By Theorem 21.4.1 there is a minimal element V in a. This V is not irreducible, otherwise we have a presentation as desired. Hence there exists a presentation V D V1 [ V2 with algebraic K-sets Vi which are strictly smaller than V . By definition, both V1 and V2 have a presentation as desired, and hence V has one, too, which gives a contradiction. Hence, a D ;. Now suppose that V D V1 [    [ Vr D W1 [    [ Ws be two presentations of the desired form. For each Vi we have a presentation Vi D .Vi \ W1 / [    [ .Vi \ Ws /. Each Vi \ Wj is a K-algebraic set (see Theorem 21.2.4). Since Vi is irreducible, we get that there is a Wj with Vi D Vi \ Wj , that is, Vi  Wj . Analogously, for this Wj there is a Vk with Wj  Vk . Altogether Vi  Wj  Vk . But Vp ª Vq if p ¤ q. Hence, from Vi  Wj  Vk we get i D k and therefore Vi D Wj , that means, for each Vi there is a Wj with Vi D Wj . Analogously, for each Wk there is a Vl with Wk D Vl . This proves the theorem.

Section 21.5 Applications and Consequences of Hilbert’s Theorems

319

Example 21.5.5. 1. Let M D ¹ghº  RŒx; y with g.x/ D x 2 C y 2  1 and 2 f .x/ D x C y 2  2. Then N .M / D V D V1 [ V2 where V1 D N .g/ and V2 D N .f /, and V is not irreducible. 2. Let M D ¹f º  RŒx; y with f .x; y/ D xy  1; f is irreducible in RŒx; y, therefore the ideal .f / is a prime ideal in RŒx; y. Hence V D N .f / is irreducible. Definition 21.5.6. Let V be an algebraic K-set in C n . The residue class ring KŒV  D KŒx1 ; : : : ; xn =I.V / is called the (affine) coordinate ring of V . KŒV  can be identified with the ring of all those functions V ! C which are given by polynomials from KŒx1 ; : : : ; xn . As a homomorphic image of KŒx1 ; : : : ; xn , we get that KŒV  can be described in the form KŒV  D KŒ˛1 ; : : : ; ˛n ; therefore a Kalgebra of the form KŒ˛1 ; : : : ; ˛n  is often called an affine K-algebra. If the algebraic K-set V in C n is irreducible – we can call V now an (affine) K-variety in C n – then KŒV  is an integral domain with an identity because I.V / is then a prime ideal with I.V / ¤ R by Theorem 21.4.2. The quotient field K.V / D Quot KŒV  is called the field of rational functions on the K-variety V . We note the following: 1. If C is algebraically closed then V D C n is a K-variety and K.V / is the field K.x1 ; : : : ; xn / of the rational functions in n variables over K. 2. Let the affine K-algebra A D KŒ˛1 ; : : : ; ˛n  be an integral domain with an identity 1 ¤ 0. Then A Š KŒx1 ; : : : ; xn =p for some prime ideal p ¤ KŒx1 ; : : : ; xn . Hence, if C is algebraically closed then A is isomorphic to the coordinate ring of the K-variety V D N .p/ in C n (see Hilbert’s Nullstellensatz, first form, Theorem 21.4.1). 3. If the affine K-algebra A D KŒ˛1 ; : : : ; ˛n  is an integral domain with an identity 1 ¤ 0 then we define the transcendence degree trgd.AjK/ to the transcendence degree of the field extension Quot.A/jK, that is, trgd.AjK/ D trgd.Quot.A/jK/, Quot.A/ the quotient field of A. In this sense trgd.KŒx1 ; : : : ; xn jK/ D n. Since Quot.A/ D K.˛1 ; : : : ; ˛n / we get trgd.AjK/  n by Theorem 20.3.10. 4. An arbitrary affine K-algebra KŒ˛1 ; : : : ; ˛n  is, as a homomorphic image of the polynomial ring KŒx1 ; : : : ; xn , noetherian (see Theorem 21.2.4 and Theorem 21.2.3). Example 21.5.7. Let !1 ; !2 2 C two elements which are linear independent over R. An element ! D m1 !1 C m2 !2 with m1 ; m2 2 Z, is called a period. The periods describe an abelian group  D ¹m1 !1 C m2 !2 W m1 ; m2 2 Zº Š Z ˚ Z and give a lattice in C.

320

Chapter 21 The Hilbert Basis Theorem and the Nullstellensatz

An elliptic function f (with respect to ) is a meromorphic function with period group , that is, f .z C w/ D f .z/ for all z 2 C. The Weierstrass }-function,  X  1 1 1  }.z/ D 2 C ; z .z  w/2 w 2 0¤w2

is an elliptic function. P P With g2 D 60 0¤w2 w14 and g3 D 140 0¤w2 w16 we get the differential equation } 0 .z/2 D 4}.z/3 C g2 }.z/ C g3 D 0. The set of elliptic functions is a field E, and each elliptic function is a rational function in } and } 0 (for details see, for instance, [27]). The polynomial f .t / D t 2  4s 3 C g2 s C g3 2 C.s/Œt  is irreducible over C.s/. For the corresponding algebraic C.s/-set V we get K.V / D C.s/Œt =.t 2  4s 3 C g2 s C g3 / Š E with respect to t 7! } 0 , s 7! }.

21.6

Dimensions

From now we assume that C is algebraically closed. Definition 21.6.1. (1) The dimension dim.V / of an algebraic K-set V in C n is said to be the supremum of all integers m for which there exists a strictly descending chain V0 © V1 ©    © Vm of K-varieties Vi in C n with Vi  V for all i. (2) Let A be a commutative ring with an identity 1 ¤ 0. The height h.p/ of a prime ideal p ¤ A of A is said to be the supremum of all integers m for which there exists a strictly ascending chain p0 ¨ p1 ¨    ¨ pm D p of prime ideals pi of A with pi ¤ A. The dimension (Krull dimension) dim.A/ of A is said to be the supremum of the heights of all prime ideals ¤ A in A. Theorem 21.6.2. Let V be an algebraic K-set in C n . Then dim.V / D dim.KŒV /. Proof. By Theorem 21.2.4 and Theorem 21.4.2 we have a bijective map between the K-varieties W with W  V and the prime ideals ¤ R D KŒx1 ; : : : ; xn  of R which contain I.V / (the bijective map reverses the inclusion). But these prime ideals correspond exactly with the prime ideals ¤ KŒV  of KŒV  D KŒx1 ; : : : ; xn =I.V / which gives the statement.

Section 21.6 Dimensions

321

Suppose that V is an algebraic K-set in C n and let V1 ; : : : ; Vr the irreducible components of V . Then dim.V / D max¹dim V1 ; : : : ; dim Vr º because if V is a K-variety with V 0  V then V 0 D .V 0 \ V1 / [    [ .V 0 \ Vr /. Hence, we may restrict ourselves on K-varieties V . If we consider the special case of the K-variety V D C 1 D C (recall that C is algebraically closed and, hence, especially C is infinite). Then KŒV  D KŒx, the polynomial ring KŒx in one indeterminate x. Now, KŒx is a principal ideal domain, and hence, each prime ideal ¤ KŒx is either a maximal ideal or the zero ideal ¹0º of KŒx. The only K-varieties in V D C are therefore V itself and the zero set of irreducible polynomials in KŒx. Hence, if V D C then dim.V / D dim KŒV  D 1 D trgd.KŒV jK/. Theorem 21.6.3. Let A D KŒ˛1 ; : : : ; ˛n  be an affine K-algebra and let A be also an integral domain. Let ¹0º D p0 ¨ p1 ¨    ¨ pm be a maximal strictly ascending chain of prime ideals in A (such a chain exists since A is noetherian). Then m D trgd.AjK/ D dim.A/. In other words: All maximal ideals of A have the same height, and this height is equal to the transcendence degree of A over K. Corollary 21.6.4. Let V be a K-variety in C n . Then dim.V / D trgd.KŒV jK/. We prove Theorem 21.6.3 in several steps. Lemma 21.6.5. Let R be an unique factorization domain. Then each prime ideal p with height h.p/ D 1 is a principal ideal. Proof. p ¤ ¹0º since h.p/ D 1. Hence there is an f 2 p, f ¤ 0. Since R is an unique factorization domain, f has a decomposition f D p1    ps with prime elements pi 2 R. Now, p is a prime ideal, hence some pi 2 p because f 2 p, say p1 2 p. Then we have the chain ¹0º ¨ .p1 /  p, and .p1 / is a prime ideal of R. Since h.p/ D 1 we get .p1 / D p. Lemma 21.6.6. Let R D KŒy1 ; : : : ; yr  be the polynomial ring of the r independent indeterminates y1 ; : : : ; yr over the field K (recall that R is a unique factorization domain). If p is a prime ideal in R with height h.p/ D 1 then the residue class ring RN D R=p has transcendence degree r  1 over K. Proof. By Lemma 21.6.5 we have that p D .p/ for some nonconstant polynomial p 2 KŒy1 ; : : : ; yr . Let the indeterminate y D yr occur in p, that is, degy .p/  1, the degree in y. If f is a multiple of p then also degy .f /  1. Hence, p \ KŒy1 ; : : : ; yr  ¤ ¹0º. Therefore the residue class mapping R ! RN D KŒyN1 ; : : : ; yNr  induces an isomorphism KŒy1 ; : : : ; yr1  ! KŒyN1 ; : : : ; yNr1  of the subring KŒy1 ; : : : ; yr1 , that is, yN1 ; : : : ; yNr1 are algebraically independent over K. On the other side p.yN1 ; : : : ;

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Chapter 21 The Hilbert Basis Theorem and the Nullstellensatz

yNr1 ; yNr / D 0 is a nontrivial algebraic relation for yNr over K.yN1 ; : : : ; yNr1 /. Hence, N altogether trgd.RjK/ D trgd.K.yN1 ; : : : ; yNr /jK/ D r  1 by Theorem 20.3.9.

Before we describe the last technical lemma we need some preparatory theoretical material. Let R; A be integral domains (with identity 1 ¤ 0) and let AjR be a ring extension. We first consider only R. (1) A subset S  R n ¹0º is called a multiplicative subset of R if 1 2 S for the identity 1 of R, and if s; t 2 S then also st 2 S. .x; s/  .y; t / W, xt  ys D 0 defines an equivalence relation on M D R  S. Let xs be the equivalence class of .x; s/ and S 1 R the set of all equivalence classes. We call xs a fraction. If we add and multiply fractions as usual we get that S 1 R becomes an integral domain; it is called the ring of fractions of R with respect to S. If, especially, S D R n ¹0º then S 1 R D Quot.R/, the quotient field of R. Now, back to the general situation. i W R ! S 1 R, i.r/ D 1r , defines an embedding of R into S 1 R. Hence, we may consider R as a subring of S 1 R. For each s 2 S  R n ¹0º we have that i.s/ is an unit in S 1 R, that is, i.s/ is invertible, and that each element of S 1 R has the form i.s/1 i.r/ with r 2 R, s 2 S . Therefore S 1 R is uniquely determined up to isomorphisms; and we have the following universal property: If  W R ! R0 is a ring homomorphism (of integral domains) such that .s/ is invertible for each s 2 S then there exist exactly one ring homomorphism  W S 1 R ! R0 with  ı i D . If a G R is an ideal in a then we write S 1 a for the ideal in S 1 R generated by i.a/. S 1 a is the set of all elements of the form as with a 2 a and s 2 S; further S 1 a D .1/ , a \ S ¤ ;. Vice versa, if A G S 1 R is an ideal in S 1 R then we denote the ideal i 1 .A/ G R with A \ R, too. An ideal a G R is of the form a D i 1 .A/ if and only if there is no s 2 S such that its image in R=a under the canonical map R ! R=a is a proper zero divisor in R=a. Under the mapping P ! P \ R and p 7! S 1 p the prime ideals in S 1 R correspond exactly to the prime ideals in R which do not contain an element of S. We now identify R with i.R/. (2) Now, let p G R be a prime ideal in R. Then S D R n p is multiplicative. In this case we write Rp instead of S 1 R and call Rp the quotient ring of R with respect to p or the localization of R of p. Put m D pRp D S 1 p. Then 1 … m. Each element of Rp =m is a unit in Rp and vice versa. In other words: Each ideal a ¤ .1/ in Rp is contained in m, or equivalently, m is the only maximal ideal in Rp . A commutative ring with an identity 1 ¤ 0, which has exactly one maximal ideal, is called a local ring. Hence Rp is a local ring. From part (1) we get further: the prime ideals of the local ring Rp correspond bijectively to the prime ideals of R which are contained in p.

323

Section 21.6 Dimensions

(3) Now we consider our ring extension AjR as above. Let q be a prime ideal in R. Claim: If qA \ R D q then there exists a prime ideal Q G A with Q \ R D q (and vice versa). Proof of the claim: If S D R n q then qA \ S D ;. Hence qS 1 A is a proper ideal in S 1 A and hence contained in a maximal ideal m in S 1 A, here qS 1 A is the ideal in S 1 A which is generated by q. Define Q D m \ A; Q is a prime ideal in A, and Q \ R D q by part (1) because Q \ S D ; where S D R n q. (4) Now let AjR be an integral extension (A; R integral domains as above). Assume that R is integrally closed in its quotient field K. Let P G A be a prime ideal in A and p D P \ R. Claim: If q G R is a prime ideal in A with q  p then qAp \ R D q. Proof of the claim: An arbitrary ˇ 2 qAp has the form ˇ D ˛s with ˛ 2 qA, qA the ideal in A generated by q, and s 2 S D A n p. An integral equation for ˛ 2 qA over K is given a form ˛ n C an1 ˛ n1 C    C a0 D 0 with ai 2 q. This can be seen as follows: we have certainly a form ˛ D b1 ˛1 C    C bm ˛m with bi 2 q and ˛i 2 A. The subring A0 D RŒ˛1 ; : : : ; ˛m  is, as an R-module, finitely generated, and ˛A0  qA0 . Now, ai 2 q follows with the same type of arguments as in the proof of Theorem 20.2.4. Now, in addition, let ˇ 2 R. Then, for s D ˇ˛ , we have an equation sn C

an1 n1 a0 C  C n D 0 s ˇ ˇ

2 R. over K. But s is integral over R, and, hence, all an1 ˇi We are now prepared to prove the last preliminary lemma which we need for the proof of Theorem 21.6.3. Lemma 21.6.7 (Krull’s going up lemma). Let AjR be an integral ring extension of integral domains and let R be integrally closed in its quotient field. Let p and q be prime ideals in R with q  p. Further let P be a prime ideal in A with P \ R D p. Then there exists a prime ideal Q in A with Q \ R D q and Q  P. Proof. It is enough to show that there exists a prime ideal Q in Ap with Q \ R D q. This can be seen from the preceding preparations. By part (1) and (2) such a Q has the form Q D Q0 Ap with a prime ideal Q0 in A with Q0  P and Q \ A D Q0 . It follows q D Q0 \ R  P \ R D p. And the existence of such a Q follows from parts (3) and (4). Proof of Theorem 21.6.3. Let first be m D 0. Then ¹0º is a maximal ideal in A and, hence, A D KŒ˛1 ; : : : ; ˛n  a field. By Corollary 20.3.11 then AjK is algebraic and, hence, trgd.AjK/ D 0. So, Theorem 21.3.3 holds for m D 0. Now, let m  1. We use Noether’s normalization theorem. A has a polynomial ring R D KŒy1 ; : : : ; yr  of the r independent indeterminates y1 ; : : : ; yr as a subring,

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Chapter 21 The Hilbert Basis Theorem and the Nullstellensatz

and AjR is an integral extension. As a polynomial ring over K the ring R is a unique factorization domain and hence, certainly, algebraically closed (in its quotient field). Now, let ¹0º D P0 ¨ P1 ¨    ¨ Pm

(1)

be a maximal strictly ascending chain of prime ideals in A. If we intersect with R we get a chain ¹0º D p0  p1      pm

(2)

of prime ideals pi D Pi \R of R. Since AjR is integral, the chain (2) is also a strictly ascending chain. This follows from Krull’s going up lemma (Lemma 21.6.7) because if pi D pj then Pi D Pj . If Pm is a maximal ideal in A then also pm is a maximal ideal in R because AjR is integral (consider A=Pm and use Theorem 17.2.21). If the chain (1) is maximal and strictly then also the chain (2). Now, let the chain (1) be maximal and strictly. If we pass to the residue class rings N1  AN D A=P1 and RN D R=p1 then we get the chains of prime ideals ¹0º D P N N P2      Pm and ¹0º D pN 1  pN 2      pN m for the affine K-algebras AN and N respectively, but with a 1 less length. By induction, we may assume that already R, N N trgd.AjK/ D m  1 D trgd.RjK/. On the other side, by construction we have trgd.AjK/ D trgd.RjK/ D r. To prove Theorem 21.3.3 finally, we have to show that N r D m. If we compare both equations then r D m follows if trgd.RjK/ D r  1. But this holds by Lemma 21.6.6. Theorem 21.6.8. Let V be a K-variety in C n . Then dim.V / D n  1 if and only if V D .f / for some irreducible polynomial f 2 KŒx1 ; : : : ; xn . Proof. (1) Let V be a K-variety in C n with dim.V / D n  1. The corresponding ideal (in the sense of Theorem 21.2.4) is by Theorem 21.4.2 a prime ideal p in KŒx1 ; : : : ; xn . By Theorem 21.3.3 and Corollary 21.3.4 we get h.p/ D 1 for the height of p because dim.V / D n  1 (see also Theorem 21.3.2). Since KŒx1 ; : : : ; xn  is a unique factorization domain we get that p D .f / is a principal ideal by Lemma 21.6.5. (2) Now let f 2 KŒx1 ; : : : ; xn  be irreducible. We have to show that V D N .f / has dimension n  1. For that, by Theorem 21.6.3, we have to show that the prime ideal p D .f / has the height h.p/ D 1. Assume that this is not the case. Then there exists a prime ideal q ¤ p with ¹0º ¤ q  p. Choose g 2 q, g ¤ 0. Let g D uf e1 2e2    rer be its prime factorization in KŒx1 ; : : : ; xn . Now g 2 q and f … q because q ¤ p. Hence, there is a i in q ¨ p D .f / which is impossible. Therefore h.p/ D 1.

Section 21.7 Exercises

21.7

325

Exercises

1. Let A D KŒa1 ; : : : ; an  and C jK a field extension with C algebraically closed. Show that there is a K-algebra homomorphism KŒa1 ; : : : ; an  ! C . 2. Let KŒx1 ; : : : ; xn  be the polynomial ring of the n independent indeterminates x1 ; : : : ; xn over the algebraically closed field K. The maximal ideals of KŒx1 ; : : : ; xn  are exactly the ideals of the form m.˛/ D .x1  ˛1 ; x2  ˛2 ; : : : ; xn  ˛n / with ˛ D .˛1 ; : : : ; ˛n / 2 K n . p 3. The nil radical 0 of A D KŒa1 ; : : : ; an  corresponds with the Jacobson radical of A, that is, the intersection of all maximal ideals of A. 4. Let R be a commutative ring with 1 ¤ 0. If each prime ideal of R is finitely generated then R is noetherian. 5. Prove the theoretical preparations for Krull’s going up lemma in detail. 6. Let KŒx1 ; : : : ; xn  be the polynomial ring of the n independent indeterminates x1 ; : : : ; xn . For each ideal a of KŒx1 ; : : : ; xn  there exists a natural number m with the following property: if f 2 KŒx1 ; : : : ; xn  vanishes on the zero set of a then f m 2 a. 7. Let K be a field with char K ¤ 2 and a; b 2 K ? . We consider the polynomial f .x; y/ D ax 2 C by 2  1 2 KŒx; y, the polynomial ring of the independent indeterminates x and y. Let C be the algebraic closure of K.x/ and ˇ 2 C with f .x; ˇ/ D 0. Show that: (i) f is irreducible over the algebraic closure C0 of K (in C ). (ii) trgd.K.x; ˇ/jK/ D 1, ŒK.x; ˇ/ W K.x/ D 2, and K is algebraically closed in K.x; ˇ/.

Chapter 22

Algebraic Cryptography

22.1

Basic Cryptography

As we have mentioned, much of mathematics has been algebraicized, that is uses the methods and techniques of abstract algebra. Throughout this book we have looked at various applications of the algebraic ideas. Many of these were to other areas of mathematics, such as the insolvability of the quintic. In this final chapter we move in a slightly different direction and look at applications of algebra to cryptography. This has become increasingly important because of the extensive use of cryptography and cryptosystems in modern commerce and communications. We first give a brief introduction to general cryptography and its history. Cryptography refers to the science and/or art of sending and receiving coded messages. Coding and hidden ciphering is an old endeavor used by governments and militaries and between private individuals from ancient times. Recently it has become even more prominent because of the necessity of sending secure and private information, such as credit card numbers, over essentially open communication systems. Traditionally cryptography is the science and or art of devising and implementing secret codes or cryptosystems. Cryptanalysis is the science and or art of breaking cryptosystems while cryptology refers to the whole field of cryptography plus cryptanalysis. In most modern literature cryptography is used synonymously with cryptology. Theoretically cryptography uses mathematics, computer science and engineering. A cryptosystem or code is an algorithm to change a plain message, called the plaintext message, into a coded message, called the ciphertext message. In general both the plaintext message (uncoded message) and the ciphertext message (coded message) are written in some N letter alphabet which is usually the same for both plaintext and code. The method of coding or the encoding algorithm is then a transformation of the N letters. The most common way to perform this transformation is to consider the N letters as N integers modulo N and then perform a number theoretical function on them. Therefore most encoding algorithms use modular arithmetic and hence cryptography is closely tied to number theory. The subject is very broad, and as mentioned above, very current, due to the need for publically viewed but coded messages. There are many references to the subject. The book by Koblitz [60] gives an outstanding introduction to the interaction between number theory and cryptography. It also includes many references to other sources. The book by Stinson [68] describes the whole area.

327

Section 22.1 Basic Cryptography

Modern cryptography is usually separated into classical cryptography also called symmetric key cryptography and public key cryptography. In the former, both the encoding and decoding algorithms are supposedly known only to the sender and receiver, usually referred to as Bob and Alice. In the latter, the encryption method is public knowledge but only the receiver knows how to decode. The message that one wants to send is written in plaintext and then converted into code. The coded message is written in ciphertext. The plaintext message and ciphertext message are written in some alphabets that are usually the same. The process of putting the plaintext message into code is called enciphering or encryption while the reverse process is called deciphering or decryption. Encryption algorithms break the plaintext and ciphertext message into message units. These are single letters or pairs of letters or more generally k-vectors of letters. The transformations are done on these message units and the encryption algorithm is a mapping from the set of plaintext message units to the set of ciphertext message units. Putting this into a mathematical formulation we let P D set of all plaintext message units C D set of all ciphertext message units: The encryption algorithm is then the application of an invertible function f W P ! C: The function f is the encryption map. The inverse f 1 W C ! P is the decryption or deciphering map. The triple ¹P ; C ; f º, consisting of a set of plaintext message units, a set of cipertext message units and an encryption map, is called a cryptosystem. Breaking a code is called cryptanalysis. An attempt to break a code is called an attack. Most cryptanalysis depends on a statistical frequency analysis of the plaintext language used (see exercises). Cryptanalysis depends also on a knowledge of the form of the code, that is, the type of cryptosystem used. We now give some examples of cryptosystems and cryptanalysis. Example 22.1.1. The simplest type of encryption algorithm is a permutation cipher. Here the letters of the plaintext alphabet are permuted and the plaintext message is sent in the permuted letters. Mathematically if the alphabet has N letters and is a permutation on 1; : : : ; N , the letter i in each message unit is replaced by .i /. For example suppose the plaintext language is English and the plaintext word is BOB and

328

Chapter 22 Algebraic Cryptography

the permutation algorithm is a b c d e f g h i j k l m b c d f g h j k l n o p r n o p q r s t u v w x y z s t v w x a e i z m q y u then BOB ! CSC. Example 22.1.2. A very straightforward example of a permutation encryption algorithm is a shift algorithm. Here we consider the plaintext alphabet as the integers 0; 1; : : : ; N  1 mod N . We choose a fixed integer k and the encryption algorithm is f Wm!mCk

mod N:

This is often known as a Caesar code after Julius Caesar who supposedly invented it. It was used by the Union Army during the American Civil War. For example if both the plaintext and ciphertext alphabets were English and each message unit was a single letter then N D 26. Suppose k D 5 and we wish to send the message ATTACK. If a D 0 then ATTACK is the numerical sequence 0; 20; 20; 0; 2; 11. The encoded message would then be FZZFIP. Any permutation encryption algorithm which goes letter to letter is very simple to attack using a statistical analysis. If enough messages are intercepted and the plaintext language is guessed then a frequency analysis of the letters will suffice to crack the code. For example in the English language the three most commonly occurring letters are E, T and A with a frequency of occurrence of approximately 13% and 9% and 8% respectively. By examining the frequency of occurrences of letters in the ciphertext the letters corresponding to E, T and A can be uncovered. Example 22.1.3. A variation on the Caesar code is the Vignère code. Here message units are considered as k-vectors of integers mod N from an N letter alphabet. Let B D .b1 ; : : : ; bk / be a fixed k-vector in Zkn . The Vignère code then takes a message unit .a1 ; : : : ; ak / ! .a1 C b1 ; : : : ; ak C bk / mod N: From a cryptanalysis point of view a Vignère code is no more secure than a Caesar code and is susceptible to the same type of statistical attack. The Alberti Code is a polyalphabetic cipher and can be often be used to thwart a statistical frequency attack. We describe it in the next example. Example 22.1.4. Suppose we have an N letter alphabet. We then form an N  N matrix P where each row and column is a distinct permutation of the plaintext alphabet. Hence P is a permutation matrix on the integers 0; : : : ; N  1. Bob and Alice

329

Section 22.1 Basic Cryptography

decide on a keyword. The keyword is placed above the plaintext message and the intersection of the keyword letter and plaintext letter below it will determine which cipher alphabet to use. We will make this precise with an 9 letter alphabet A, B, C, D, E, O, S, T, U. Here for simplicity we will assume that each row is just a shift of the previous row, but any permutation can be used. Key Letters A B C D E O S T U a A a b c d e o s t u l B b c d e o s t u a p C c d e o s t u a b h D d e o s t u a b c a E e o s t u a b c d b O o s t u a b c d e e S s t u a b c d e o t T t u a b c d e o s s U u a b c d e o s t. Suppose the plaintext message is STAB DOC and Bob and Alice have chosen the keyword BET. We place the keyword repeatedly over the message B E T B E T B S T A B D O C: To encode we look at B which lies over S. The intersection of the B key letter and the S alphabet is a t so we encrypt the S with T. The next key letter is E which lies over T. The intersection of the E keyletter with the T alphabet is c. Continuing in this manner and ignoring the space we get the encryption STAB DOC ! TCTCTDD: Example 22.1.5. A final example, which is not number theory based, is the so-called Beale Cipher. This has a very interesting history which is related in the popular book Archimedes Revenge by Paul Hoffman (see [56]). Here letters are encrypted by numbering the first letters of each word in some document like the Declaration of Independence or the Bible. There will then be several choices for each letter and a Beale cipher is quite difficult to attack. Until relatively recent times cryptography was mainly concerned with message confidentiality – that is sending secret messages so that interceptors or eavesdroppers cannot decipher them. The discipline was primarily used in military and espionage situations. This changed with the vast amount of confidential data that had to be transmitted over public airways so the field has expanded to many different types of cryptographic techniques such as digital signatures and message authentications.

330

Chapter 22 Algebraic Cryptography

Cryptography and encryption does have a long and celebrated history. In the Bible, in the book of Jeremiah, they use what is called an Atabash Code. In this code the letters of the alphabet – Hebrew in the Bible but can be used with any alphabet – are permuted first to last. That is, in the Latin alphabet, Z would go to A and so on. The Kabbalists and the Kabbala believe that the Bible – written in Hebrew where each letter also stands for a number – is a code from heaven. They have devised elaborate ways to decode it. This idea has seeped into popular culture where the book “The Bible Code” became a bestseller. In his military campaigns Julius Caesar would send out coded messages. His method, which we looked at in the last section, is now known as a Caesar code. It is a shift cipher. That is each letter is shifted a certain amount to the right. A shift cipher is a special case of an affine cipher that will be elaborated upon in the next section. The Caesar code was resurrected and used during the American Civil War. Coded messages produced by most of the historical methods reveal statistical information about the plaintext. This could be used in most cases to break the codes. The discovery of frequency analysis was done by the Arab mathematician Al-Kindi in the ninth century and the basic classical substitution ciphers became more or less easily breakable. About 1470 Leon Alberti developed a method to thwart statistical analysis. His innovation was to use a polyalphabetic cipher where different parts of the message are encrypted with different alphabets. We looked at an example of an Alberti code in this section. A different way to thwart statistical attacks is to use blank and neutral letters, that is meaningless letters within the message. Mary, Queen of Scots, used a random permutation cipher with neutrals in it, where a neutral was a random meaningless symbol. Unfortunately for her, her messages were decoded and she was beheaded. There have been various physical devices and aids used to create codes. Prior to the widespread use of the computer the most famous cryptographic aid was the Enigma machine developed and used by the German military during the Second World War. This was a rotor machine using a polyalphabetic cipher. An early version was broken by Polish cryptographers early in the war so a larger system was built that was considered unbreakable. British cryptographers led by Alan Turing broke this and British knowledge of German secrets had a great effect on the latter part of the war. The development of digital computers allowed for the development of much more complicated cryptosystems. Further this allowed for the encryption using anything that can be placed in binary formats whereas historical cryptosystems could only be rendered using language texts. This has revolutionized cryptography. In 1976 Diffie and Hellman developed the first usable public key exchange protocol. This allowed for the transmission of secret data over open airways. A year later Rivest, Adelman and Shamir, developed the RSA algorithm, a second public key protocol. There are now many and we will discuss them later. In 1997 it became known that public key cryptography had been developed earlier by James Ellis working for British

331

Section 22.2 Encryption and Number Theory

Intelligence and that both the Diffie–Hellman and RSA protocols had been developed earlier by Malcom Williamson and Clifford Cocks respectively.

22.2

Encryption and Number Theory

Here we describe some basic number theoretically derived cryptosystems. In applying a cryptosystem to an N letter alphabet we consider the letters as integers mod N . The encryption algorithms then apply number theoretic functions and use modular arithmetic on these integers. One example of this was the shift, or Caesar cipher, described in the last section. In this encryption method a fixed integer k is chosen and the encryption map is given f Wm!mCk

mod N:

The shift algorithm is a special case of an affine cipher. Recall that an affine map on a ring R is a function f .x/ D ax C b with a; b; x 2 R. We apply such a map to the ring of integers modulo n, that is, R D Zn , as the encryption map. Specifically again suppose we have an N letter alphabet and we consider the letters as the integers 0; 1; : : : ; N  1 mod N , that is in the ring ZN . We choose integers a; b 2 ZN with .a; N / D 1 and b ¤ 0. a; b are called the keys of the cryptosystem . The encryption map is then given by f W m ! am C b mod N: Example 22.2.1. Using an affine cipher with the English language and keys a D 3, b D 5 encode the message EAT AT JOE’S. Ignore spaces and punctuation. The numerical sequence for the message ignoring the spaces and punctuation is 4; 0; 19; 0; 19; 9; 14; 4; 18: Applying the map f .m/ D 3m C 5 mod 26 we get 17; 5; 62; 5; 62; 32; 47; 17; 59 ! 17; 5; 10; 5; 10; 6; 21; 17; 7: Now rewriting these as letters we get EAT AT JOE’S ! RFKFKGVRH: Since .a; N / D 1 the integer a has a multiplicative inverse a1 mod N . The decryption map for an affine cipher with keys a; b is then f 1 W m ! a1 .m  b/ mod N: Since an affine cipher, as given above, goes letter to letter it is easy to attack using a statistical frequency approach. Further if an attacker can determine two letters and

332

Chapter 22 Algebraic Cryptography

knows that it is an affine cipher the keys can be determined and the code broken. To give better security it is preferable to use k-vectors of letters as message units. The form then of an affine cipher becomes f W v ! Av C B where here v and B are k-vectors from ZkN and A is an invertible k  k matrix with entries from the ring ZN . The computations are then done modulo N . Since v is a k-vector and A is a k  k matrix the matrix product Av produces another k-vector from ZkN . Adding the k-vector B again produces a k-vector so the ciphertext message unit is again a k-vector. The keys for this affine cryptosystem are the enciphering matrix A and the shift vector B. The matrix A is chosen to be invertible over ZN (equivalent to the determinant of A being a unit in the ring ZN ) so the decryption map is given by v ! A1 .v  B/: Here A1 is the matrix inverse over ZN and v is a k-vector. The enciphering matrix A and the shift vector B are now the keys of the cryptosystem. A statistical frequency attack on such a cryptosystem requires knowledge, within a given language, of the statistical frequency of k-strings of letters. This is more difficult to determine than the statistical frequency of single letters. As for a letter to letter affine cipher, if k C 1 message units, where k is the message block length, are discovered, then the code can be broken. Example 22.2.2. Using an affine cipher with message units of length 2 in the English language and keys   5 1 AD ; B D .5; 3/ 8 7 encode the message EAT AT JOE’S. Again ignore spaces and punctuation. Message units of length 2, that is 2-vectors of letters are called digraphs. We first must place the plaintext message in terms of these message units. The numerical sequence for the message EAT AT JOE’S ignoring the spaces and punctuation is as before 4; 0; 19; 0; 19; 9; 14; 4; 18: Therefore the message units are .4; 0/; .19; 0/; .19; 9/; .14; 4/; .18; 18/ repeating the last letter to end the message. The enciphering matrix A has determinant 1 which is a unit mod 26 and hence is invertible so it is a valid key.

333

Section 22.2 Encryption and Number Theory

Now we must apply the map f .v/ D Av CB mod 26 to each digraph. For example              4 5 1 4 5 20 5 25 A CB D C D C D : 0 8 7 0 3 32 3 9 Doing this to the other message units we obtain .25; 9/; .22; 25/; .5; 10/; .1; 13/; .9; 13/: Now rewriting these as digraphs of letters we get (Z, J), (W, Z), (F, K), (B, N), (J, N): Therefore the coded message is EAT AT JOE’S ! ZJWZFKBNJN: Example 22.2.3. Suppose we receive the message ZJWZFKBNJN and we wish to decode it. We know that an affine cipher with message units of length 2 in the English language and keys   5 1 AD ; B D .5; 3/ 8 7 is being used. The decryption map is given by v ! A1 .v  B/: so we must find the inverse matrix for A. For a 2  2 invertible matrix 

a b c d

1

1 D ad  bc

a

b c d



 d b : c a

Therefore in this case recalling that multiplication is mod 26     5 1 7 1 AD H) A1 D : 8 7 8 5 The message ZJWZFKBNJN in terms of message units is .25; 9/; .22; 25/; .5; 10/; .1; 13/; .9; 13/: We apply the decryption map to each digraph. For example         20 7 1 25 5 B D  D .4; 0/: A1 6 8 5 9 3



we have

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Chapter 22 Algebraic Cryptography

Doing this to each we obtain .4; 0/; .19; 0/; .19; 9/; .14; 4/; .18; 18/ and rewriting in terms of letters (E, A), (T, A), (T, J), (O, E), (S, S): This gives us ZJWZFKBNJN ! EATATJOESS: Modern cryptography is done via a computer. Hence all messages both plaintext and ciphertext are actually presented as binary strings. Important in this regard is the concept of a hash function. A cryptographic hash function is a deterministic function h W S ! ¹0; 1ºn that returns for each arbitrary block of data, called a message, a fixed size bit string. It should have the property that a change in the data will change the hash value. The hash value is called the digest. An ideal cryptographic hash function has the following properties: (1) It is easy to compute the hash value for any given message. (2) It is infeasible to find a message that has a given hash value (preimage resistant). (3) It is infeasible to modify a message without changing its hash. (4) It is infeasible to find two different messages with the same hash (collision resistant). A cryptographic hash function can serve as a digital signature. Hash functions can also be used with encryption. Suppose that Bob and Alice want to communicate openly. They have exchanged a secret key K that supposedly only they know. Let fK be an encryption function or encryption algorithm based on the key K. Alice wants to send the message m to Bob and m is given as a binary bit string. Alice sends to Bob fK .m/ ˚ h.K/ where ˚ is addition modulo 2. Bob knows the key K and hence its hash value h.K/. He now computes fK .m/ ˚ h.K/ ˚ h.K/:

Section 22.3 Public Key Cryptography

335

Since addition modulo 2 has order 2 we have fK .m/ ˚ h.K/ ˚ h.K/ D fk .M /: Bob now applies the decryption algorithm fK1 to decode the message. Alice could have just as easily sent fK .m/ ˚ K. However sending the hash has two benefits. Usually the hash is shorter than the key and from the properties of hash functions it gives another level of security. As we will see, tying the secret key to the actual encryption in this manner is the basis for the El-Gamal and elliptic curve cryptographic methods. The encryption algorithm fK is usually a symmetric key encryption so that anyone knowing K can encrypt and decrypt easily. However it should be resistant to plaintext-ciphertext attacks. That is if an attacker gains some knowledge of a piece of plaintext together with the corresponding ciphertext it should not compromise the whole system. The encryption algorithm can either be a block cipher or a stream cipher. In the former, blocks of fixed length k are transformed into blocks of fixed length n and there is a method to ties the encrypted blocks together. In the latter, a stream cipher, bits are transformed one by one into new bit strings by some procedure. In 2001 the National Institute of Standards and Technology adopted a block cipher now called AES for Advanced Encryption System as the industry standard for a symmetric key encryption. Although not universally used it is the most widely used. This block cipher was a standardization of the Rijnadel cipher named after its inventor Rijmen and Daeman. AES replaced DES or Digital Encryption System which had been the standard. Parts of DES were found to be insecure. AES proceeds with several rounds of encrypting blocks and then mixing blocks. The mathematics in AES is done over the finite field GF.28 /.

22.3

Public Key Cryptography

Presently there are many instances where secure information must be sent over open communication lines. These include for example banking and financial transactions, purchasing items via credit cards over the Internet and similar things. This led to the development of public key cryptography. Roughly, in classical cryptography only the sender and receiver know the encoding and decoding methods. Further it is a feature of such cryptosystems, such as the ones that we’ve looked at, that if the encrypting method is known then the decryption can be carried out. In public key cryptography the encryption method is public knowledge but only the receiver knows how to decode. More precisely in a classical cryptosystem once the encrypting algorithm is known the decryption algorithm can be implemented in approximately the same order of magnitude of time. In a public key cryptosystem, developed first by Diffie and

336

Chapter 22 Algebraic Cryptography

Hellman, the decryption algorithm is much more difficult to implement. This difficulty depends on the type of computing machinery used and as computers get better, new and more secure public key cryptosystems become necessary. The basic idea in a public key cryptosystem is to have a one-way function or trapdoor function. That is a function which is easy to implement but very hard to invert. Hence it becomes simple to encrypt a message but very hard, unless you know the inverse, to decrypt. The standard model for public key systems is the following. Alice wants to send a message to Bob. The encrypting map fA for Alice is public knowledge as well as the encrypting map fB for Bob. On the other hand the decryption algorithms fA1 and fB1 are secret and known only to Alice and Bob respectively. Let P be the message Alice wants to send to Bob. She sends fB fA1 .P /. To decode Bob applies first fB1 , which only he knows. This gives him fB1 .fB fA1 .P // D fA1 .P /. He then looks up fA which is publically available and applies this fA .fA1 .P // D P to obtain the message. Why not just send fB .P /. Bob is the only one who can decode this. The idea is authentication, that is being certain from Bob’s point of view that the message really came from Alice. Suppose P is Alice’s verification; signature, social security number etc. If Bob receives fB .P / it could be sent by anyone since fB is public. On the other hand since only Alice supposedly knows fA1 getting a reasonable message from fA .fB1 fB fA1 .P // would verify that it is from Alice. Applying fB1 alone should result in nonsense. Getting a reasonable one way function can be a formidable task. The most widely used (at present) public key systems are based on difficult to invert number theoretic functions. The original public key system was developed by Diffie and Hellman in 1976. It was followed closely by a second public key system developed by Rivest, Shamir and Adeelman known as the RSA system. Although at present there are many different public key systems in use most are variations of these original two. The variations are attempts to make the systems more secure. We will discuss four such systems.

22.3.1 The Diffie–Hellman Protocol Diffie and Hellman in 1976 developed the original public key idea using the discrete log problem. In modular arithmetic it is easy to raise an element to a power but difficult to determine, given an element, if it is a power of another element. Specifically if G is a finite group, such as the cyclic multiplicative group of Zp where p is a prime, and h D gk for some k then the discrete log of h to the base g is any integer t with h D gt . The rough form of the Diffie–Hellman public key system is as follows. Bob and Alice will use a classical cryptosystem based on a key k with 1 < k < q  1 where q is a prime. It is the key k that Alice must share with Bob. Let g be a multiplicative

Section 22.3 Public Key Cryptography

337

generator of Z?q the multiplicative group of Zq . The generator g is public. It is known that this group is cyclic if q is a prime. Alice chooses an a 2 Zq with 1 < a < q  1. She makes public ga . Bob chooses a b 2 Z?q and makes public g b . The secret key is gab . Both Bob and Alice, but presumably none else, can discover this key. Alice knows her secret power a and the value g b is public from Bob. Hence she can compute the key gab D .g b /a . The analogous situation holds for Bob. An attacker however only knows g a and g b and g. Unless the attacker can solve the discrete log problem the key exchange is secure. Given q; g; g a ; g b the problem of determining the secret key gab is called the Diffie–Hellman problem. At present the only known solution is to solve the discrete log problem which appears to be very hard. In choosing the prime q and the generator g it is assumed that the prime q is very large so that the order of g is very large. There are algorithms to solve the discrete log problem is q is too small. One attack on the Diffie–Hellman key exchange is a man in the middle attack. Since the basic protocol involves no authentication an attacker can pretend to be Bob and get information from Alice and then pretend to be Alice and get information from Bob. In this way the attacker could get the secret shared key. To prevent this, digital signatures are often used (see [60] for a discussion of these). The decision Diffie–Hellman problem is given a prime q and g a mod q, g b mod q and g c mod q determine if gc D g ab . In 1997 it became known that the ideas of public key cryptography were developed by British Intelligence Services prior to Diffie and Hellman.

22.3.2 The RSA Algorithm In 1977 Rivest, Adelman and Shamir developed the RSA algorithm, which is presently (in several variations) the most widely used public key cryptosystem. It is based on the difficulty of factoring large integers and in particular on the fact that it is easier to test for primality than to factor very large integers. In basic form the RSA algorithm works as follows. Alice chooses two large primes pA ; qA and an integer eA relatively prime to .pA qA / D .pA  1/.qA  1/ where  is the Euler phi-function. It is assumed that these integers are chosen randomly to minimize attack. Primality tests arise in the following manner. Alice first randomly chooses a large odd integer m and tests it for primality. If its prime it is used. If not, she tests m C 2, m C 4 and so on until she gets her first prime pA . She then repeats the process to get qA . Similarly she chooses another odd integer m and tests until she gets an eA relatively prime to .pA qA /. The primes she chooses should be quite large. Originally RSA used primes of approximately 100 decimal digits, but as computing and attack have become more sophisticated, larger primes have had to be utilized. Presently keys with 400 decimal digits are not uncommon. Once Alice has obtained pA ; qA ; eA she lets nA D pA qA and computes dA , the multiplicative inverse of eA modulo .nA /. That is dA satisfies eA dA  1 mod .pA  1/.qA  1/. She makes

338

Chapter 22 Algebraic Cryptography

public the enciphering key KA D .nA ; eA / and the encryption algorithm known to all is fA .P / D P eA mod nA where P 2 ZnA is a message unit. It can be shown that if .eA ; .pA  1/.qA  1// D 1 and eA dA  1 mod .pA  1/.qA  1/ then P eA dA  P mod nA (see exercises). Therefore the decryption algorithm is fA1 .C / D C da

mod nA :

Notice then that fA1 .fA .P // D P eA dA  P mod nA so it is the inverse. Now Bob makes the same type of choices to obtain pB ; qB ; eB . He lets nB D pB qB and makes public his key KB D .nB ; eB /. If Alice wants to send a message to Bob that can be authenticated to be from Alice she sends fB .fA1 .P //. An attack then requires factoring nA or nB which is much more difficult than obtaining the primes pA ; qA ; pB ; qB . In practice suppose there is an N letter alphabet which is to be used for both plaintext and ciphertext. The plaintext message is to consist of k vectors of letters and the ciphertext message of l vectors of letters with k < l. Each of the k plaintext letters in a message unit P are then considered as integers mod N and the whole plaintext message is considered as a k digit integer written to the base N (see example below). The transformed message is then written as an l digit integer mod N and then the digits are then considered integers mod N from which encrypted letters are found. To ensure that the range of plaintext messages and ciphertext messages are the same k < l are chosen so that N k < nU < N l for each user U, that is nU D pU qU . In this case any plaintext message P is an integer less than N k considered as an element of ZnU . Since nU < N l the image under the power transformation corresponds to an l digit integer written to the base N and hence to an l letter block. We give an example with relatively small primes. In real world applications the primes would be chosen to have over a hundred digits and the computations and choices must be done using good computing machinery. Example 22.3.1. Suppose N D 26, k D 2 and l D 3. Suppose further that Alice chooses pA D 29, qA D 41, eA D 13. Here nA D 29  41 D 1189 so she makes public the key KA D .1189; 13/. She then computes the multiplicative inverse dA of 13 mod 1120 D 28  40. Now suppose we want to send her the message TABU. Since k D 2 the message units in plaintext are 2 vectors of letters so we separate the message into TA BU. We show how to send TA. First the numerical sequence for the letters TA mod 26 is (19,0). We then use these as the digits of a 2-digit number to the base 26. Hence TA D 19  26 C 0  1 D 494:

339

Section 22.3 Public Key Cryptography

We now compute the power transformation using her eA D 13 to evaluate f .19; 0/ D 49413

mod 1189:

This is evaluated as 320. Now we write 320 to the base 26. By our choices of k; l this can be written with a maximum of 3 digits to this base. Then 320 D 0  262 C 12  26 C 8: The letters in the encoded message then correspond to .0; 12; 8/ and therefore the encryption of TA is AMI. To decode the message Alice knows dA and applies the inverse transformation. Since we have assumed that k < l this seems to restrict the direction in which messages can be sent. In practice to allow messages to go between any two users the following is done. Suppose Alice is sending an authenticated message to Bob. The keys kA D .nA ; eA /; kB D .nB ; eB / are public. If nA < nB Alice sends fB fA1 .P /. On the other hand if nA > nB she sends fA1 fB .P /. There have been attacks on RSA for special types of primes so care must be chosen in choosing the primes. The computations and choices used in real world implementations of the RSA algorithm must be done with computers. Similarly, attacks on RSA are done via computers. As computing machinery gets stronger and factoring algorithms get faster, RSA becomes less secure and larger and larger primes must be used. In order to combat this, other public key methods are in various stages of ongoing development. RSA and Diffie–Hellman and many related public key cryptosystems use properties in abelian groups. In recent years a great deal of work has been done to encrypt and decrypt using certain nonabelian groups such as linear groups or braid groups. We will discuss these later in the chapter.

22.3.3 The El-Gamal Protocol The El-Gamal cryptosystem is a method to use the Diffie–Hellman key exchange method to do encryption. The method works as follows and uses the fact that hash functions can also be used with encryption. Suppose that Bob and Alice want to communicate openly. They have exchanged a secret key K that supposedly only they know. Let fK be an encryption function or encryption algorithm based on the key K. Alice wants to send the message m to Bob and m is given as a binary bit string. Alice sends to Bob fK .m/ ˚ h.K/ where ˚ is addition modulo 2. Bob knows the key K and hence its hash value h.K/. He now computes fK .m/ ˚ h.K/ ˚ h.K/:

340

Chapter 22 Algebraic Cryptography

Since addition modulo 2 has order 2 we have fK .m/ ˚ h.K/ ˚ h.K/ D fk .M /: Bob now applies the decryption algorithm fK1 to decode the message. Hence if K is a publicly exchanged secret key and fK is a cryptosystem based on K then the above format allows an encryption algorithm to go with the key exchange. The El-Gamal system does this with the Diffie–Hellman key exchange protocol. Suppose that Bob and Alice want to communicate openly. Alice chooses a prime q and a generator g of the multiplicative group Zq . q should be large enough to thwart the known discrete logarithm algorithms. Alice then chooses an integer a with 1 < a < q  1. She then computes A D ga

mod q:

Her public key is then .q; g; A/. Bob wants to send a message M to Alice. He first encrypts the message an integer m mod q. For Bob to now send an encrypted message m to Alice he chooses a random integer b with 1 < b < q  2 and compute B D gb

mod q:

Bob then sends to Alice the integer c D Ab m mod q that is Bob encrypts the whole message by multiplying it by the Diffie–Hellman shared key. The complete El-Gamal ciphertext is then the pair .B; c/. How does Alice decode the message. Given the message m she knows how to reconstruct the plaintext message M so she must recover the mod q integer m. As in the Diffie–Hellman key exchange she can compute the shared key Ab D B a . She can then divide c by this Diffie–Hellman key g ab to obtain m. To avoid having to find the inverse of B a mod q which can be difficult she computes the exponent x D p  1  a. The inverse is then B x mod q. For each new El-Gamal encryption a new exponent b is chosen so that there is a random component of El-Gamal which improves the security. Breaking the El-Gamal system is as difficult as breaking the Diffie–Hellman protocol and hence is based on the difficulty of the discrete log problem. However the El-Gamal has the advantage that the choice of primes is random. As mentioned the primes should be chosen large enough to not be susceptible to known discrete log algorithms. Presently the primes should be of binary length at least 512. c D Ab m

mod q:

341

Section 22.3 Public Key Cryptography

22.3.4 Elliptic Curves and Elliptic Curve Methods A very powerful approach which has had wide ranging applications in cryptography is to use elliptic curves. If F is a field of characteristic not equal to 2 or 3 then an elliptic curve over F is the locus of points .x; y/ 2 F  F satisfying the equation y 2 D x 3 C ax C b

with 4a3 C 27b 2 ¤ 0:

We denote by 0 a single point at infinity and let E.F / D ¹.x; y/ 2 F  F W y 2 D x 3 C ax C bº [ ¹0º: The important thing about elliptic curves from the viewpoint of cryptography is that a group structure can be placed on E.F /. In particular we define the operation C on E.F / by 1. 0 C P D P for any point P 2 E.F /. 2. If P D .x; y/ then P D .x; y/ and 0 D 0. 3. P C .P / D 0 for any point P 2 E.F /. 4. If P1 D .x1 ; y1 /; P2 D .x2 ; y2 / with P1 ¤ P2 then P1 C P2 D .x3 ; y3 /

with x3 D m2  .x1 C x2 /;

and

y3 D m.x3  x1 /  y1

mD

y2  y 1 x2  x 1

if x2 ¤ x1

mD

3x12 C a 2y1

if x2 D x1 :

and

This operation has a very nice geometric interpretation if F D R the real numbers. It is known as the chord and tangent method. If P1 ¤ P2 are two points on the curve then the line through P1 ; P2 intersects the curve at another point P3 . If we reflect P3 through the x-axis we get P1 C P2 . If P2 D P2 we take the tangent line at P1 . With this operation E.F / becomes an abelian group (due to Cassels) whose structure can be worked out. Theorem 22.3.2. E.F / together with the operations defined above forms an abelian group. In F is a finite field of order p k then E.F / is either cyclic or has the structure E.F / D Zm1  Zm2 with m1 jm2 and m1 j.p k  1/.

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Chapter 22 Algebraic Cryptography

A comprehensive description and discussion of elliptic curve methods can be found in Crandall and Pomerance [52]. The groups of elliptic curves can be used for cryptography as developed by Koblitz and others. If q is a prime and a; b 2 Zq then we can form the elliptic curve E.p W a; b/ and the corresponding elliptic curve abelian group. In this group the Diffie– Hellman key exchange protocol and the corresponding El-Gamal encryption system can be implemented. Care must be taken that the discrete log problem in E.qI a; b/ is difficult. The curve is then called a cryptographically secure elliptic curve. Elliptic curve public-key cryptosystems are at present the most important commutative alternatives to the use of the RSA algorithm. There are several reasons for this. They are more efficient in many cases than RSA and keys in elliptic curve systems are much smaller than keys in RSA. It is felt that it is important to have good workable alternatives to RSA in the event that factoring algorithms become strong enough to compromise RSA encryption.

22.4

Noncommutative Group based Cryptography

The public key cryptosystems and public key exchange protocols that we have discussed, such as the RSA algorithm, Diffie–Hellman, El-Gamal and elliptic curve methods are number theory based and hence depend on the structure of abelian groups. Although there have been no overall successful attacks on the standard methods there is a feeling that the strength of computing machinery has made these techniques theoretically susceptible to attack. As a result of this, there has been a recent active line of research to develop cryptosystems and key exchange protocols using noncommutative cryptographic platforms. This line of investigation has been given the broad title of noncommutative algebraic cryptography. Since most of the cryptographic platforms are groups this is also known as group based cryptography. The book by Myasnikov, Shpilrain and Ushakov [63] provides an overview of group based cryptographic methods tied to complexity theory. Up to this point the main sources for noncommutative cryptographic platforms has been nonabelian groups. In cryptosystems based on these objects algebraic properties of the platforms are used prominently in both devising cryptosystems and in cryptanalysis. In particular the nonsolvability of certain algorithmic problems in finitely presented groups, such as the conjugator search problem, has been crucial in encryption and decryption. The main sources for nonabelian groups are combinatorial group theory and linear group theory. Braid group cryptography (see [53]), where encryption is done within the classical braid groups, is one prominent example. The one way functions in braid group systems are based on the difficulty of solving group theoretic decision problems such as the conjugacy problem and conjugator search problem. Although braid group cryptography had initial spectacular success, various potential attacks

Section 22.4 Noncommutative Group based Cryptography

343

have been identified. Borovik, Myasnikov, Shpilrain [49] and others have studied the statistical aspects of these attacks and have identified what are termed black holes in the platform groups outside of which present cryptographic problems. Baumslag. Fine and Xu in [46] and [69] suggested potential cryptosystems using a combination of combinatorial group theory and linear groups and a general schema for the these types of cryptosystems was given. In [47] a public key version of this schema using the classical modular group as a platform was presented. A cryptosystem using the extended modular group SL2 .Z/ was developed by Yamamura [70] but was subsequently shown to have loopholes [67]. In [47] attacks based on these loopholes were closed. The extension of the cryptographic ideas to noncommutative platforms involves the following ideas, (1) General Algebraic Techniques for Developing Cryptosystems (2) Potential Algebraic Platforms (Specific Groups, Rings, etc.) for implementing the Techniques (3) Cryptanalysis and Security Analysis of the Resulting Systems. The main source for noncommutative platforms are nonabelian groups and the main method for handling nonabelian groups in cryptography is combinatorial group theory which we discussed in detail in Chapter 14. The basic idea in using combinatorial group theory for cryptography is that elements of groups can be expressed as words in some alphabet. If there is an easy method to rewrite group elements in terms of these words and further the technique used in this rewriting process can be supplied by a secret key then a cryptosystem can be created. One of the earliest descriptions of a free group cryptosystem was in a paper by W. Magnus in the early 1970s [61]. Recall that the classical modular group M is M D PSL2 .Z/. Hence M consists of the 2  2 projective integral matrices: ²   ³ a b M D ˙ W ad  bc D 1; a; b; c; d 2 Z : c d Equivalently M can be considered as the set of integral linear fractional transformations with determinant 1: z0 D

az C b ; cz C d

ad  bc D 1;

a; b; c; d 2 Z:

Magnus proved the following theorem. Theorem 22.4.1 ([46]).  1 ˙ 1

The matrices    1 1 C 4t 2 2t ; ; ˙ 2t 1 2

t D 1; 2; 3; : : :

344

Chapter 22 Algebraic Cryptography

freely generate a free subgroup F of infinite index in M . Further distinct elements of F have distinct first columns. Since the entries in the generating matrices are positive we can do the following. Choose a set T1 ; : : : ; Tn of projective matrices from the set above with n large enough to encode a desired plaintext alphabet A. Any message would be encoded by a word W .T1 ; : : : ; Tn / with nonnegative exponents. This represents an element g of F . The two elements in the first column determine W and therefore g. Receiving W then determines the message uniquely. The idea of using the difficulty of group theory decision problems in infinite nonabelian groups was first developed by Magyarik and Wagner in 1985. They devised a public key protocol based on the difficulty of the solution of the word problem (see Chapter 14). Although this was a seminal idea their basic cryptosystem was really unworkable and not secure in the form they presented. Wagner and Magyarik outlined a conceptual public key cryptosystem based on the hardness of the word problem for finitely presented groups. At the same time, they gave a specific example of such a system. González Vasco and Steinwandt proved that their approach is vulnerable to so-called reaction attacks. In particular, for the proposed instance it is possible to retrieve the private key just by watching the performance of a legitimate recipient. The general scheme of the Wagner and Magyarik public-key cryptosystemis as follows. Let X be a finite set of generators, and let R and S be finite sets of relators such that the group G0 D hXI R [ S i has an easy word problem. That is the word problem can be solved in polynomial time while the G D hX I Ri has a hard word problem (see Chapter 14 for terminology). Choose two words W0 and W1 which are not equivalent in G0 (and hence not equivalent in G). The public key is the presentation hXI Ri and the chosen words W0 and W1 . To encrypt a single bit 2 ¹0; 1º, pick Wi and transform it into a ciphertext word W by repeatedly and randomly applying Tietze transformations to the presentation hX I Ri. To decrypt a word W , run the algorithm for the word problem of G0 in order to decide which of Wi W 1 is equivalent to the empty word for the presentation hXI R [ Si The private key is the set S. Actually, this is not sufficient and Wagner and Magyarik are not clear on this point. The public key should be a deterministic polynomial-time algorithm for the word problem of G0 D hX I R [ Si. Just knowing S does not automatically and explicitly give us an efficient algorithm (even if such an algorithm exists).

Section 22.4 Noncommutative Group based Cryptography

345

22.4.1 Free Group Cryptosystems The simplest example of a nonabelian group based cryptosystem is perhaps a free group cryptosystem. This can be described in the following manner. Consider a free group F on free generators x1 ; : : : ; xr . Then each element g in F has a unique expression as a word W .x1 ; : : : ; xr /. Let W1 ; : : : ; Wk with Wi D Wi .x1 ; : : : ; xr / be a set of words in the generators x1 ; : : : ; xr of the free group F . At the most basic level, to construct a cryptosystem, suppose that we have a plaintext alphabet A. For example, suppose that A D ¹a; b; : : :º are the symbols needed to construct meaningful messages in English. To encrypt, use a substitution ciphertext A ! ¹W1 ; : : : ; Wk º: That is a 7! W1 ;

b 7! W2 ; : : : :

Then given a word W .a; b; : : :/ in the plaintext alphabet form the free group word W .W1 ; W2 ; : : :/. This represents an element g in F . Send out g as the secret message. In order to implement this scheme we need a concrete representation of g and then for decryption a way to rewrite g back in terms of W1 ; : : : ; Wk . This concrete representation is the idea behind homomorphic cryptosystems. The decryption algorithm in a free group cryptosystem then depends on the Reidemeister–Schreier rewriting process. As described in Chapter 14 this is a method to rewrite elements of a subgroup of a free group in terms of the generators of that subgroup. Recall that roughly it works as follows. Assume that W1 ; : : : ; Wk are free generators for some subgroup H of a free group F on ¹x1 ; : : : ; xn º. Each Wi is then a reduced word in the generators ¹x1 ; : : : ; xn º. A Schreier transversal for H is a set ¹h1 ; : : : ; h t ; : : :º of (left) coset representatives for H in F of a special form (see Chapter 14). Any subgroup of a free group has a Schreier transversal. The Reidemeister– Schreier process allows one to construct a set of generators W1 ; : : : ; Wk for H by using a Schreier transversal. Further given the Schreier transversal from which the set of generators for H was constructed, the Reidemeister–Schreier rewriting process allows us to algorithmically rewrite an element of H . Given such an element expressed as a word W D W .x1 ; : : : ; xr / in the generators of F this algorithm rewrites W as a word W ? .W1 ; : : : ; Wk / in the generators of H . The knowledge of a Schreier transversal and the use of Reidemeister–Schreier rewriting facilitates the decoding process in the free group case but is not essential. Given a known set of generators for a subgroup the Stallings Folding Method to develop a subgroup graph can also be utilized to rewrite in terms of the given generators. The paper by Kapovich and Myasnikov [58] is now a standard reference for this method in free groups. At present there is an ongoing study of the complexity of Reidemeister–Schreier being done by Baumslag, Brukhov, Fine and Troeger.

346

Chapter 22 Algebraic Cryptography

Pure free group cryptosystems are subject to various attacks and can be broken easily. However a public key free group cryptosystem using a free group representation in the Modular group was developed by Baumslag, Fine and Xu [46, 47]. The most successful attacks on free group cryptosystems are called length based attacks. Here an attacker multiplies a word in ciphertext by a generator to get a shorter word which could possibly be decoded. Baumslag, Fine and Xu in [46] described the following general encryption scheme using free group cryptography. A further enhancement was discussed in the paper [47]. We start with a finitely presented group G D hXjRi where X D ¹x1 ; : : : ; xn º and a faithful representation W G ! G: G can be any one of several different kinds of objects – linear group, permutation group, power series ring etc. We assume that there is an algorithm to re-express an element of .G/ in G in terms of the generators of G. That is if g D W .x1 ; : : : ; xn ; : : :/ 2 G where W is a word in the these generators and we are given .g/ 2 G we can algorithmically find g and its expression as the word W .x1 ; : : : ; xn /. Once we have G we assume that we have two free subgroups K; H with H  K  G: We assume that we have fixed Schreier transversals for K in G and for H in K both of which are held in secret by the communicating parties Bob and Alice. Now based on the fixed Schreier transversals we have sets of Schreier generators constructed from the Reidemeister–Schreier process for K and for H . k1 ; : : : ; km ; : : :

for K

h1 ; : : : ; h t ; : : :

for H:

and Notice that the generators for K will be given as words in x1 ; : : : ; xn the generators of G while the generators for H will be given as words in the generators k1 ; k2 ; : : : for K. We note further that H and K may coincide and that H and K need not in general be free but only have a unique set of normal forms so that the representation of an element in terms of the given Schreier generators is unique. We will encode within H , or more precisely within .H /. We assume that the number of generators for H is larger than the set of characters within our plaintext

347

Section 22.4 Noncommutative Group based Cryptography

alphabet. Let A D ¹a; b; c; : : :º be our plaintext alphabet. At the simplest level we choose a starting point i, within the generators of H , and encode a 7! hi ;

b 7! hi C1 ; : : :

etc.

Suppose that Bob wants to communicate the message W .a; b; c; : : :/ to Alice where W is a word in the plaintext alphabet. Recall that both Bob and Alice know the various Schreier transversals which are kept secret between them. Bob then encodes W .hi ; hiC1 ; : : :/ and computes in G the element W . .hi /; .hi C1 /; : : :/ which he sends to Alice. This is sent as a matrix if G is a linear group or as a permutation if G is a permutation group and so on. Alice uses the algorithm for G relative to G to rewrite W . .hi /; .hi C1 /; : : :/ as a word W ? .x1 ; : : : ; xn / in the generators of G. She then uses the Schreier transversal for K in G to rewrite using the Reidemeister–Schreier process W ? as a word W ?? .k1 ; : : : ; ks ; : : :/ in the generators of K. Since K is free or has unique normal forms this expression for the element of K is unique. Once she has the word written in the generators of K she uses the transversal for H in K to rewrite again, using the Reidemeister–Schreier process, in terms of the generators for H . She then has a word W ??? .hi ; hiC1 ; : : :/ and using hi 7! a; hi C1 7! b; : : : decodes the message. In actual implementation an additional random noise factor is added. In [46] and [47] an implementation of this process was presented that used for the base group G the classical modular group M D PSL2 .Z/. Further it was a polyalphabetic cipher which was secure. The system in the modular group M was presented as follows. A list of finitely generated free subgroups H1 ; : : : ; Hm of M is public and presented by their systems of generators (presented as matrices). In a full practical implementation it is assumed that m is large. For each Hi we have a Schreier transversal h1;i ; : : : ; h t.i/;i and a corresponding ordered set of generators W1;i ; : : : ; Wm.i /;i constructed from the Schreier transversal by the Reidemeister–Schreier process. It is assumed that each m.i/ l where l is the size of the plaintext alphabet, that is each subgroup has many more generators than the size of the plaintext alphabet. Although Bob and Alice know these subgroups in terms of free group generators what is made public are generating systems given in terms of matrices. The subgroups on this list and their corresponding Schreier transversals can be chosen in a variety of ways. For example the commutator subgroup of the Modular group is free of rank 2 and some of the subgroups Hi can be determined from homomorphisms of this subgroup onto a set of finite groups.

348

Chapter 22 Algebraic Cryptography

Suppose that Bob wants to send a message to Alice. Bob first chooses three integers .m; q; t / where m D choice of the subgroup Hm q D starting point among the generators of Hm for the substitution of the plaintext alphabet t D size of the message unit: We clarify the meanings of q and t . Once Bob chooses m, to further clarify the meaning of q, he makes the substitution a 7! Wm;q ;

b 7! Wm;qC1 ; : : : :

Again the assumption is that m.i/ l so that starting almost anywhere in the sequence of generators of Hm will allow this substitution. The message unit size t is the number of coded letters that Bob will place into each coded integral matrix. Once Bob has made the choices .m; q; t / he takes his plaintext message W .a; b; : : :/ and groups blocks of t letters. He then makes the given substitution above to form the corresponding matrices in the Modular group; T1 ; : : : ; Ts : We now introduce a random noise factor. After forming T1 ; : : : ; Ts , Bob then multiplies on the right each Ti by a random matrix in M say RTi (different for each Ti ). The only restriction on this random matrix RTi is that there is no free cancellation in forming the product Ti RTi . This can be easily checked and ensures that the freely reduced form for Ti RTi is just the concatenation of the expressions for Ti and RTi . Next he sends Alice the integral key .m; q; t / by some public key method (RSA, Anshel–Anshel–Goldfeld etc.). He then sends the message as s random matrices T1 RT1 ; T2 RT2 ; : : : ; Ts RTs : Hence what is actually being sent out are not elements of the chosen subgroup Hm but rather elements of random right cosets of Hm in M . The purpose of sending coset elements is two-fold. The first is to hinder any geometric attack by masking the subgroup. The second is that it makes the resulting words in the modular group generators longer – effectively hindering a brute force attack. To decode the message Alice first uses public key decryption to obtain the integral keys .m; q; t /. She then knows the subgroup Hm , the ciphertext substitution from the generators of Hm and how many letters t each matrix encodes. She next uses the algorithms described in Section 14.4 to express each Ti RTi in terms of the free group generators of M say WTi .y1 ; : : : ; yn /. She has knowledge of the Schreier

Section 22.5 Ko–Lee and Anshel–Anshel–Goldfeld Methods

349

transversal, which is held secretly by Bob and Alice, so now uses the Reidemeister– Schreier rewriting process to start expressing this freely reduced word in terms of the generators of Hm . The Reidemeister–Schreier rewriting is done letter by letter from left to right (see Chapter 14). Hence when she reaches t of the free generators she stops. Notice that the string that she is rewriting is longer than what she needs to rewrite in order to decode as a result of the random polynomial RTi . This is due to the fact that she is actually rewriting not an element of the subgroup but an element in a right coset. This presents a further difficulty to an attacker. Since these are random right cosets it makes it difficult to pick up statistical patterns in the generators even if more than one message is intercepted. In practice the subgroups should be changed with each message. The initial key .m; q; t / is changed frequently. Hence as mentioned above this method becomes a type of polyalphabetic cipher. Polyalphabetic ciphers have historically been very difficult to decode. A further variation of this method using a formal power series ring in noncommuting variables over a field was described in [43]. There have been many cryptosystems based on the difficulty of solving hard group theoretic problems. The book by Myasnikov, Shpilrain and Ushakov [63] describes many of these in detail.

22.5

Ko–Lee and Anshel–Anshel–Goldfeld Methods

After the initial attempt by Wagner and Magyarik to develop a cryptosystem based on a hard group theoretic problem there have been many developments using nonabelian groups in cryptography. Among the first were the cryptographic schemes of Anshel, Anshel and Goldfeld [42] and Ko and Lee [59]. Both sets of authors, at about the same time, proposed using nonabelian groups and combinatorial group theory for public key exchange. The security of these systems depended on the difficulty of solving certain “hard” group theoretic problems. The methods of both Anshel–Anshel–Goldfeld and Ko–Lee can be considered as group theoretic analogs of the number theory based Diffie–Hellman method. The basic underlying idea is the following. If G is a group and g; h 2 G we let gh denote the conjugate of g by h, that is g h D h1 gh. The simple observation is that this behaves like ordinary exponentiation in that .gh1 /h2 D g h1 h2 . From this straightforward idea one can exactly mimic the Diffie–Hellman protocol within a nonabelian group. Both the Anshel–Anshel–Goldfeld protocol and the Ko–Lee protocol start with a platform group G given by a group presentation. A major assumption in both protocols is that the elements of G have nice unique normal forms that are easy to compute for given group elements. However it is further assumed that given normal forms for x; y 2 G the normal form for the product xy does not reveal x or y.

350

Chapter 22 Algebraic Cryptography

22.5.1 The Ko–Lee Protocol Ko and Lee [59] developed a public key exchange system that is a direct translation of the Diffie–Hellman protocol to a nonabelian group theoretic setting. Its security is based on the difficulty of the conjugacy problem. We again assume that the platform group has nice unique normal forms that are easy to compute given a group element but hard to recover the group element. Recall again that gh means the conjugate of g by h. In the Ko–Lee protocol, Alice and Bob choose commuting subgroups A and B of the platform group G. A is Alice’s subgroup while Bob’s subgroup is B and these are secret. Now they completely mimic the classical Diffie–Hellman technique. There is a public element g 2 G, Alice chooses a random secret element a 2 A and makes public g a . Bob chooses a random secret element b 2 B and makes public gb . The secret shared key is g ab . Notice that ab D ba since the subgroups commute. It follows then, that .g a /b D gab D gba D .g b /a just as if these were exponents. Hence both Bob and Alice can determine the common secret. The difficulty is in the difficulty of the conjugacy problem. The conjugacy problem for a group G, or more precisely for a group presentation for G is given g; h 2 G to determine algorithmically if they are conjugates. As with the conjugator search problem it is known that the conjugacy is undecidable in general but there are groups where it is but hard. These groups then become the target platform groups for the Ko–Lee protocol. As with the Anshel–Anshel–Goldfeld protocol, Ko and Lee suggest the use of the braid groups. As with the standard Diffie–Hellman key exchange protocol using number theory the Ko–Lee protocol can be changed to an encryption system via the El-Gamal method. Their are several different variants of noncommutative El-Gamal systems.

22.5.2 The Anshel–Anshel–Goldfeld Protocol We now describe the Anshel–Anshel–Goldfeld public key exchange protocol. Let G be the platform group given by a finite presentation and with the assumptions on normal forms as described above. Alice and Bob want to communicate a shared secret. First, Alice and Bob choose random finitely generated subgroups of G by giving a set of generators for each. A D ¹a1 ; : : : ; an º;

B D ¹b1 ; : : : ; bm º

and make them public. The subgroup A is Alice’s subgroup while the subgroup B is Bob’s subgroup. Alice chooses a secret group word a D W .a1 ; : : : ; an / in her subgroup while Bob chooses a secret group word b D V .b1 ; : : : ; bm / in his subgroup. For an element g 2 G we let NF.g/ denote the normal form for g. Alice knows her secret word a and knows the generators bi of Bob’s subgroup. She makes public the normal forms

Section 22.6 Platform Groups and Braid Group Cryptography

351

of the conjugates NF.bia /;

i D 1; : : : ; m:

Bob knows his secret word b and the generators ai of Alice’s subgroup and makes public the normal forms of the conjugates NF.ajb /;

j D 1; : : : ; n:

The common shared secret is the commutator Œa; b D a1 b 1 ab D a1 ab D .b a /1 b: Notice that Alice knows ab since she knows a in terms of generators ai of her subgroup and she knows the conjugates by b since Bob has made the conjugates of the generators of A by b public. Since Alice knows ab she knows Œa; b D a1 ab . In an analogous manner Bob knows Œa; b D .b a /1 b. An attacker would have to know the corresponding conjugator, that is the element that conjugates each of the generators. Given elements g; h in a group G where it is known that g k D k 1 gk D h the conjugator search problem is to determine the conjugator k. It is known that this problem is undecidable in general, that is there are groups where the conjugator cannot be determined algorithmically. On the other hand there are groups where the conjugator search problem is solvable but “difficult”, that is the complexity of solving the conjugator search problem is hard. Such groups become the ideal platform groups for the Anshel–Anshel–Goldfeld protocol. The security in this system is then in the difficulty of the conjugator search problem. Anshel, Anshel and Goldfeld suggested the Braid Groups as potential platforms, use for example B80 with 12 or more generators in the subgroups. Their suggestion and that of Ko and Lee led to development of braid group cryptography. There have been various attacks on the Braid group system. However some have been handled by changing the parameters. In general the ideas remain valid despite the attacks. The Anshel–Anshel–Goldfeld key exchange can be developed into a cryptosystem again by the El-Gamal method. There have been many other public key exchange protocols developed using nonabelian groups. A large number of them are described in the book of Myasnikov, Shpilrain and Ushakov [63]. The authors of that book themselves have developed many of these methods. They use different “hard” group theoretic decision problems and many have been broken. On the other hand the security of many of them is still open and they perhaps can be used as viable alternatives to commutative methods.

22.6

Platform Groups and Braid Group Cryptography

Given a group based encryption scheme, such as Ko–Lee or Anshel–Anshel–Goldfeld a platform group is a group G in which the encryption is to take place. In general, platform groups for the noncommutative protocols that we have discussed require certain

352

Chapter 22 Algebraic Cryptography

properties. The first is the existence of a normal form for elements in the group. Normal forms provide an effective method of disguising elements. Without this, one can determine a secret key simply by inspection of group elements. Further if N.x/; N.y/ are the normal forms for x; y respectively then it should difficult to determine N.x/ and N.y/ from N.xy/. The existence of a normal form in a group implies that the group has solvable word problem, which is essential for these protocols. For purposes of practicality, the group also needs an efficiently computable normal form, which ensures an efficiently solvable word problem. In addition to the platform group having normal form, ideally, it would also be large enough so that a brute force search for the secret key is infeasible. Currently, there are many potential platform groups that have been suggested. The following are some of the proposals. We refer to [63] for a discussion of many of these. 

Braid groups (Ko–Lee, Anshel–Anshel–Goldfeld),



Thompson Groups (Shpilrain–Ushakov) [65],



Polycyclic Groups (Eick–Kahrobaei) [54],



Linear Groups (Baumslag–Fine–Xu) [46, 47],



Free metabelian Groups (Shpilrain–Zapata) [66],



Artin Groups (Shpilrain–Zapata) [66],



Grigorchuk Groups (Petrides) [64],



Groups of Matrices (Grigoriev–Ponomarenko) [55],



Surface Braid Groups (Camps) [51].

Most of these are discussed in detail in [63]. As platform groups for their respective protocols, both Ko–Lee and Anshel– Anshel–Goldfeld suggested the braid groups Bn (see [50]). The groups in this class of groups possess the desired properties for the key exchange and key transport protocols; they have nice presentations with solvable word problems and conjugacy problems; the solution to the conjugacy and conjugator search problem is “hard”; there are several possibilities for normal forms for element and they have many choices for large commuting subgroups. Initially the braid groups were considered so ideal as platforms that many other cryptographic applications were framed within the braid group setting. These included authentication, that is identifying over a public airwave that a message received was from the correct sender and digital signature, that is sending an encrypted message with an included authentication. There was so much enthusiasm about using these groups that the whole area of study was named braid group cryptography. A comprehensive and well-written article by Dehornoy [31] provides a detailed overview of the subject and we refer the reader to that for technical details.

Section 22.6 Platform Groups and Braid Group Cryptography

353

After the initial successes with braid group cryptographic schemes there were some surprisingly effective attacks. There were essentially three types of attacks; an attack using solutions to the conjugacy and conjugator search problems, an attack using heuristic probability within Bn and an attack based on the fact that there are faithful linear representations of each Bn (see [31]). What is most surprising is that the Anshel–Anshel–Goldfeld method was susceptible to a length based attack. In the Anshel–Anshel–Goldfeld method the parameters are the specific braid group Bn and the rank of the secret subgroups for Bob and Alice. A length based attack essentially broke the method for the initial parameters suggested by AAG. The parameters were then made larger and attacks by this method were less successful. However this led to research on why these attacks on the conjugator search problem within Bn were successful. What was discovered was that generically a random subgroup of Bn is a free group and hence length based attacks are essentially attacks on free group cryptography and therefore successful. What this indicated was that although randomness is important in cryptography in using the braid groups as platforms subgroups cannot be chosen purely randomly. Braid groups arise in several different areas of mathematics and have several equivalent formulations. We close this chapter and the book with a brief introduction to braid groups. A complete topological and algebraic description can be found in the book of Joan Birman [50]. A braid on n strings is obtained by starting with n parallel strings and intertwining them. We number the strings at each vertical position and keep track of where each individual string begins and ends. We say that two braids are equivalent if it is possible to move the strings of one of the braids in space without moving the endpoints or moving through a string and obtain the other braid. A braid with no crossings is called a trivial braid. We form a product of braids in the following manner. If u is the first braid and v is the second braid then uv is the braid formed by placing the starting points for the strings in v at the endpoints of the strings in u. The inverse of a braid is the mirror image in the horizontal plane. It is clear that if we form the product of a braid and its mirror image we get a braid equivalent to the trivial braid. With these definitions the set of all equivalence classes braids on n strings forms a group Bn . We let i denote the braid that has a single crossing from string i over string i C 1. Since a general braid is just a series of crossings it follows that Bn is generated by the set i ; i D 1; : : : ; n  1. There is an equivalent algebraic formulation of the braid group Bn . Let Fn be a free on the n generators x1 ; : : : ; xn with n > 2. Let i , i D 1; : : : ; n  1 be the automorphism of Fn given by

i W xi 7! xi C1 ; xi C1 7! xi1 C1 xi xi C1

i W xj 7! xj ;

j ¤ i; i C 1:

354

Chapter 22 Algebraic Cryptography

Then each i corresponds precisely to the basic crossings in Bn . Therefore Bn can be considered as the subgroup of Aut.Fn / generated by the automorphisms i , Artin proved [28] (see also [24]) that a finite presentation for Bn is given by Bn D h 1 ; : : : ; n1 I Œ i ; j  D 1 if ji  j j > 1; xi C1 xi xi C1 D xi xi C1 xi ; i D 1; : : : ; n  1i: This is now called the Artin presentation. The fact that Bn is contained in Aut.Fn / provides an elementary solution to the word problem in Bn since one can determine easily if an automorphism of Fn is trivial on all the generators. We note that although the braid groups Bn are linear (the Lawrence-Krammer representation is faithful (see [31]) it is known that Aut.Fn / is not linear (see [34]). From the commuting relations in the Artin presentation it is clear that each Bn has the requisite collection of commuting subgroups. The conjugacy problem for Bn was originally solved by Garside and it was assumed that it was hard in the complexity sense. Recently there has been significant research on the complexity of the solution to the conjugacy problem (see [63] and [31]). There are several possibilities for normal forms for elements of Bn . The two most commonly used are the Garside normal form and the Dehornoy handle form. These are described in [31] and [63]. For braid group cryptography one must be careful in using more than one normal form in an encryption scheme. The second may expose what the first is hiding and vice versa (see [31]). We describe first the Dehornoy handle form. Let W be a word in the generators of the braid group Bn . An xi -handle is a subword of W of the form xi V xi with  D ˙1 and where the word V does not involve xi . If V does not contain any xiC1 -handles then the xi -handle is called permitted A braid word W is obtained from a braid word W 0 by a one step handle reduction if some subword of W is a permitted xi -handle xi V xi and W 0 is obtained from W by applying the following substitutions for all letters in the xi -handle xj˙1 ! 1

if j D i

˙1  xj˙1 ! xi C1 xi xi C1

if j D i C 1

xj˙1 ! xj˙1

if j < i or j > i C 1:

W can be obtained from W 0 by an m-step handle reduction if W can be obtained from W 0 by a sequence of m one-step handle reductions. A word is handle free if it has no handles. The handle free braid words provide normal forms for the elements of Bn .

Section 22.6 Platform Groups and Braid Group Cryptography

355

Theorem 22.6.1. Let W be a braid word. Then the following holds: (1) Any sequence of handle reductions applied to W will eventually stop and produce a handle free braid word V representing the same element as W . (2) The word W represents the identity in Bn if and only if any sequence of handle reductions applied to W produces the trivial word or equivalently the handle free form of W is trivial. The handle free reduction process is very efficient and most of the time works in polynomial time on the length of the braid word to produce the handle free form. However there is no known theoretical complexity estimate (see [31]). Garside solved the conjugacy problem using a different type of normal form for Bn . Let Sn be the symmetric group on n letters and for each s 2 Sn let s be the shortest positive braid such that . s / D s. The elements S D ¹ s W s 2 Sn º  Bn are called simple elements. We order the simple elements so that s < t if there exists r 2 Sn such that t D s r . This produces a lattice structure on S. The trivial braid is the smallest element of S while the greatest element of S is the half-twist braid  D .n;n1;:::;2;1/ : The Garside left normal form of a braid a 2 Bn is a pair .p; .s1 ; : : : ; s t // where p 2 Z and s1 ; : : : ; s t is a sequence of permutations in Sn n ¹1; º satisfying for each i D 1; : : : ; t  1 1 D gcd. s 1 ; siC1 / i

where gcd. s ; t / D max¹ r W r < s and r < t º: A normal form .p; .s1 ; : : : ; s t // represents the element p s1 : : : sn :

Theorem 22.6.2. There exists an algorithm which computes the normal form of the corresponding braid for any braid word W D w.x1 ; : : : ; xn /.

356

22.7

Chapter 22 Algebraic Cryptography

Exercises

1. Show that if p; q are primes and e; d are positive integers with .e; .p1/.q1// D 1 and ed  1 mod .p  1/.q  1/ then aed  a mod pq for any integer a. (This is the basis if the decryption function used in the RSA algorithm. 2. The following table gives the approximate statistical frequency of occurrence of letters in the English language. The passage below is encrypted with a simple permutation cipher without punctuation. Use a frequency analysis to try to decode it. letter frequency letter frequency letter frequency A :082 B :015 C :028 D :043 E :127 F :022 G :020 H :061 I 070 J :002 K :008 L :040 M :024 N :067 O :075 P :019 Q :001 R :060 S :063 T :091 U :028 V :010 W :023 X :001 Y :020 Z :001 ZKIRNVMFNYVIRHZKLHRGREVRMGVTVIDSR XSSZHZHGHLMOBKLHRGREVWRERHLIHLMVZ MWRGHVOUKIRNVMFNYVIHKOZBZXIFXRZOI LOVRMMFNYVIGSVLIBZMWZIVGSVYZHRHUL IGHSHVMLGVHGSVIVZIVRMURMRGVOBNZMB KIRNVHZMWGSVBHVIEVZHYFROWRMTYOLXP HULIZOOGSVKLHRGREVRMGVTVIH 3. Encrypt the message NO MORE WAR using an affine cipher with single letters keys a D 7, b D 5. 4. Encrypt the message NO MORE WAR using an affine cipher on 2 vectors of letters and an encrypting keys   5 2 AD ; B D .3; 7/: 1 1 5. What is the decryption algorithm for the affine cipher given in the last problem. 6. How many different affine enciphering transformations are there on single letters with an N letter alphabet. 7. Let N 2 N with N  2 and n ! an C b with .a; N / D 1 is an affine cipher on an N letter alphabet. Show that if any two letters are guessed n1 ! m1 , n2 ! m2 with .n1  n2 ; N / D 1 then the code can be broken.

357

Section 22.7 Exercises

8. Let F be a free group of rank 3 with generators x; y; z. Code the English alphabet by a 7! 0, b 7! 1; : : : . Consider the free group cryptosystem given by i 7! Wi where Wi D x i y i C1 z i C2 x i C1 . Code the message EAT AT JOES with this system. 9. In the Anshel–Anshel–Goldfeld protocol verify that both Bob and Alice will know the commutator.

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Index

A abelian group, 3, 101 abelianization, 176 adjoining a root, 92 AES, 335 affine cipher, 331 affine coordinate ring, 319 algebraic closure, 74, 91, 95 algebraic extension, 69 algebraic geometry, 312 algebraic integer, 295 algebraic number field, 297 algebraic numbers, 67, 75 algebraic variety, 312 algebraically closed, 91, 94 alternating group, 166 annihilator, 270 Anshel–Anshel–Goldfeld protocol, 350 associates, 35 automorphism, 11 axiom of choice, 26 axiom of well-ordering, 26

B basis theorem for finite abelian groups, 151, 285 Betti number, 287 block cipher, 335 braid group, 353 braid group cryptography, 353

C Cardano’s formulas, 256 Cayley graph, 211 Cayley’s theorem, 127 cell complex, 209 centralizer, 182 characteristic, 15 ciphertext, 327 class equation, 183 combinatorial group theory, 192 commutative algebra, 312

commutative ring, 3 commutator, 175 composition series, 178 congruence motion, 123 conjugacy class, 181 conjugacy problem, 213 constructible number, 80 construction of a regular n-gon, 84 coset, 18, 128 cryptanalysis, 326, 327 cryptography, 326 public key, 327 symmetric key, 327 cryptology, 326 cryptosystem, 326 cyclic group, 121 cyclotomic field, 252

D decryption, 327 Dedekind domain, 51 Dehornoy handle form, 354 derived series, 176 Diffie–Hellman protocol, 336 dihedral groups, 156 dimension of an algebraic set, 320 discrete log problem, 336 divisibility, 29 division algorithm, 30 doubling the cube, 83 Dyck’s theorem, 212

E Eisenstein’s criterion, 62 El-Gamal protocol, 339 elliptic curve methods, 341 elliptic function, 320 encryption, 327 Euclid’s lemma, 21 Euclidean algorithm, 32 Euclidean domain, 45 Euclidean group, 123

364 Euclidean norm, 45 extension field, 66

F factor group, 19, 144 factor ring, 9 Feit–Thompson theorem, 189 field, 4 extension, 66 field extension, 66 algebraic, 69 by radicals, 248 degree, 67 finite, 67 finitely generated, 69 isomorphic, 67 separable, 233 simple, 69 transcendental, 69 field of fractions, 14 finite fields, 236 finite integral domains, 6 fix field, 220 free group, 193 rank, 196 free group cryptosystems, 345 free modules, 273 free product, 214 free reduction, 194 Frobenius homomorphism, 16 Fuchsian group, 201 fundamental theorem of algebra, 105, 261 fundamental theorem of arithmetic, 29 fundamental theorem of Galois theory, 221 fundamental theorem of modules, 279 fundamental theorem of symmetric polynomials, 104

G Galois extension, 233 finite, 220 Galois group, 218 Galois theory, 217 Garside normal form, 355 Gauss’ lemma, 58

Index Gaussian integers, 47 Gaussian primes, 49 Gaussian rationals, 48 general linear group, 123 group, 17, 101, 119 abelian, 3, 17, 119 center, 181 conjugate elements, 181 coset, 128 cyclic, 134 direct product, 150 finite, 17, 101, 119 finitely generated, 199 finitely presented, 199 finitely related, 199 free abelian, 287 free product, 214 generating system, 199 generators, 127, 199 homomorphism, 121 internal direct product, 150 isomorphism, 121 order, 17, 101, 119 presentation, 127, 199 relations, 127 relator, 199 simple, 168 solvable, 172 transversal, 128 group action, 180 group based cryptography, 342 group isomorphism theorem, 19, 146 group presentation, 199 group table, 120

H hash function, 334 Hilbert basis theorem, 315 Hilbert’s Nullstellensatz, 316 homomorphism group, 17 automorphism, 17 epimorphism, 17 isomorphism, 17 monomorphism, 17 ring, 11 automorphism, 11

365

Index endomorphism, 11 epimorphism, 11 isomorphism, 11 monomorphism, 11

I ideal, 7 generators, 27 maximal, 24 prime, 22 product, 23 ideals in Z, 8 index of a subgroup, 18 insolvability of the quintic, 254 integral closure, 300 integral domain, 4 integral element, 298 integral ring extension, 299 integrally closed, 300 intermediate field, 67 irreducible element, 35 isometry, 123 isomorphism problem, 213

J Jordan–Hölder theorem, 178

K K-isomorphism, 91 kernel, 19 Ko–Lee protocol, 350 Kronecker’s theorem, 91 Krull dimension, 320 Krull’s lemma, 323 Kurosh theorem, 215

L Lagrange’s theorem, 18 local ring, 322

M maximal ideal, 24 minimal polynomial, 70 modular group, 200 modular rings, 5 modular rings in Z, 11 module, 265

N Nielsen–Schreier theorem, 197 noetherian, 314 noncommutative algebraic cryptography, 343 norm, 36 normal extension, 116 normal forms, 197 normal series, 172 normal subgroup, 18, 142 normalizer, 183

O one-way function, 336

P p-group, 157 p-Sylow subgroup, 159 perfect field, 233 permutation, 17, 101 permutation cipher, 327 permutation group, 126 plaintext, 327 platform group, 351 polynomial, 42, 53 coefficients, 43, 53 constant, 43 degree, 42, 53 irreducible, 44, 54, 55 leading coefficient, 43, 53 linear, 43, 54 prime, 44, 55 primitive, 57 quadratic, 43, 54 separable, 233 zero, 42 zero of, 55 Prüfer ring, 51 prime element, 35 prime field, 14 prime ideal, 22 prime ring, 15 primitive element theorem, 245 principal ideal, 8, 27 principal ideal domain, 27 public key cryptosystem, 336 purely transcendental, 305

366

Q quotient group, 19, 144 quotient ring, 9

R R-algebra, 298 R-module, 265 cyclic, 267 direct product, 271 factor module, 268 faithful, 271 free, 273 generators, 268 quotient module, 268 torsion element, 270 unitary, 266 radical, 314 nil, 314 rational integers, 49 rational primes, 49 Reidemeister–Schreier process, 207 ring, 2 commutative, 3 finite, 3 prime, 15 trivial, 3 with identity, 3 ring extension, 298 ring isomorphism theorem, 12 ring of polynomials, 54 RSA algorithm, 337

Index subfield, 6 subgroup, 18, 101, 120 commutator, 175 conjugate, 141 cyclic, 121 derived, 175 index, 129 normal, 142 subring, 6 Sylow theorems, 160, 183 symmetric group, 17, 101, 161 symmetric polynomials, 104 symmetry, 124

T transcendence basis, 303 transcendence degree, 305, 319 transcendental extension, 69 transcendental numbers, 67, 75 transitive action, 180 transposition, 164 trapdoor function, 336 trisecting an angle, 83

U UFD, 38 unique factorization domain, 38 unit, 4, 35 unit group, 35

V vector space, 66

S

W

separable field extension, 233 separable hull, 241 separable polynomial, 233 simple extension, 69 simple group, 168 simplicial complex, 209 solvability by radicals, 248 solvable group, 172 solvable series, 172 special linear group, 123 splitting field, 100, 113 squaring the circle, 83 stabilizer, 126, 181 stream cipher, 335

Wagner–Magyarik system, 344 word, 194 cyclically reduced, 197 length, 194 reduced, 194 trivial, 194 word problem, 213

Z zero divisor, 4 Zorn’s lemma, 26